4

As discussed in the question Triangles on a Torus, the outside of a torus has positive Gaussian curvature. A triangle inscribed on the outer surface will have interior angles which add to be greater than 180 degrees.

By adjusting the geometry of the Torus, and the points which define the triangle, what is the maximum possible sum of interior angles of the triangle?

I suspect the fringe case occurs with a horn torus with vertices that lie on the axis between positive and negative curvature.

Edit: “Inside of the Torus” refers to the side of the torus facing inward “toward the donut hole.” Alternatively, it is simply the surface of the torus with negative Gaussian curvature.

Edit 2: I’m referring to one single triangle drawn anywhere on the surface of the torus, not breaking the surface into a number of smaller triangles

Arctic Char
  • 16,007
Phillip D
  • 53
  • 3
    How are you defining "triangle" and "inside"? Have you considered a triangle which is (or is close to being) the outer equator of the torus? – Henry Sep 16 '22 at 17:30
  • Henry, your suggestion that the maximum solution could have vertices which lie on the outer equator is certainly a possibility. I’m unfamiliar with how to test the hypothesis, hence asking the question. – Phillip D Sep 16 '22 at 18:10
  • You can certainly go bigger for some definitions of "inside" – Henry Sep 16 '22 at 18:29
  • 1
    I assume the edges of your triangles need not be geodesics? Please specify carefully. – Ted Shifrin Sep 16 '22 at 19:18
  • I believe the edges are geodesics, as in the shortest line between two points on a curved surface. In other words, to an ant on the surface, it’s a straight line. – Phillip D Sep 16 '22 at 20:36
  • This is a torus embedded in $\mathbb{R}^3$, rather than say $(\mathbb{R}/\mathbb{Z}) \times (\mathbb{R}/\mathbb{Z})$? – aschepler Sep 17 '22 at 13:00
  • By Gauss Bonnet thm intuitively $ A+B+C=\Omega+\pi $ ... the patch of maximum integral curvature. – Narasimham Sep 19 '22 at 06:37
  • @ Phillip D: Sorry about several changes to my answer as I first commented with intuition and later on built some logic into it. – Narasimham Sep 19 '22 at 08:14

1 Answers1

2

Consider three points on the positive Gauss curvature outer $K>0$ region. Points (T,M,B) are at top, middle and bottom in poloidal meridian principal direction respectively on a circular torus on "outside" viewing. Alternately we can also say $(T,B)$ to be points on the crown where $ K=0$ and $M$ is on outer equator, with $K>0$.

As a first guess in the curvilinear toroidal triangle:

Two geodesics helically going all around the axis of symmetry fully for $ \theta_{max}=2 \pi$ rotation clockwise MT,MB.

and,

TMB, the poloidal meridian for the third geodesic side...

may make up a maximum angle sum.

By Gauss Bonnet thm $$ \int K dS + \int k_g ds + \Sigma_i= 2 \pi$$ $$ \Omega +0 + \Sigma_i= 2 \pi$$ $$ \Omega +3 \pi -(A+B+C)=2 \pi$$ $$ (A+B+C)=\Omega + \pi $$

so that the geodesic triangle should also have maximum integral curvature $\Omega.$Accordingly $\theta$ is chosen as its maximum $ 2 \pi$ in a single cover mentioned as basis at start somewhat like in the following crude sketch.

At $T,B$ geodesics run along circumferential lines so from Clairaut's Law we have

$$\sin \alpha=\frac{r_T}{r_M} $$

The internal angles sum up to $\pi+ 2\alpha$ where $ 2\alpha $ is angle at wedge M.

$$ R=r_T,\; \frac{R+r}{2}=r_M,\text { when}\; R\to\infty, \alpha= 30^{\circ},\text{angle sum} =180^{\circ}+ 2 \cdot 30^{\circ}= 240^{\circ}.$$

enter image description here

Narasimham
  • 40,495
  • Your answer was thorough and adequate, thank you. I'm wondering if another case is a possible solution, would you mind testing it? Consider a triangle with vertices at B,B and T(as shown in your diagram), so that it wraps around the entire torus. do you think it would have similar behavior? Or perhaps it's not worth testing, I'm not familiar with the math to intuit the difference. – Phillip D Sep 20 '22 at 01:28
  • Glad it was of help. Can test it. (B,T are points on either side of $K=0$ crown line). But could not understand what the triangle vertices are.. do you mean all around area between two close poloidal lines with a common point? Actually I also thought of it but being a curved quadrilateral, dropped it. – Narasimham Sep 20 '22 at 11:58
  • Narasimham, sorry I didn't formulate my question very well. I'm imagining a triangle with one edge made up of the bottom crown line, and a point on the top crown line. Such a triangle would have two vertices at the same point on the bottom crown line, and one vertex on the top crown line 180 degrees around the major axis of the torus. – Phillip D Sep 23 '22 at 16:00
  • Trying to understand. Is it like the one sketched at left? but that triangle has 3 vertices alright but 4 sides ! – Narasimham Sep 23 '22 at 18:49
  • Not quite. Let’s say the torus has major axis R and minor axis r. The crown line has length 2piR. So a imagine a triangle of width 2piR and height pi*r is wrapped around the outside of the torus. – Phillip D Sep 25 '22 at 00:42