This problem was inspired by having read some cool theorems about tiling higher-dimensional things with simplices (n-dimensional polytopes that are isometric to the convex hull of n+1 points in general position in n-space — the natural generalization of a triangle). Definitions: * An n-simplex is "acute" if all of its dihedral angles (there are n(n+1)/2 of them, one for every pair of (n-1)-dimensional faces) is acute. * A "flat" n-manifold is one such that every point has a neighborhood that is isometric with some open ball in n-dimensional Euclidean space. * An "acute triangulation" of a flat manifold X is a way of expressing X as the union of acute simplices with disjoint interiors such that any (n-1)-face of any simplex belongs to exactly two simplices. * The cubical n-torus is R^n / Z^n, the result of identifying opposite (n-1)-faces of the unit n-cube. Theorem: -------- There is an acute triangulation of the 3-cube having 1370 tetrahedra. (This is apparently the fewest known.) Theorem: -------- There exists no acute triangulation of n-space for n >= 5. Theorem: -------- There exists no acute triangulation of the cubical n-torus for n >= 4. (It is unknown whether there exists any acute triangulation of 4-space.) Theorem: -------- If there exists an acute triangulation of 4-space, then the supremum of its set of dihedral angles must be pi/2. –Dan

On Jan 31, 2017, at 6:16 AM, Veit Elser <ve10@cornell.edu> wrote:

By Euler the average degree of a vertex is 6, and the number of triangles is twice the number of vertices. I can construct a tiling from 3 vertices, for 6 triangles (first slice Z^2 on square diagonals to make strips, then subdivide strips). Ruling out a tiling with just one vertex looks doable, but the two vertex case might be tedious …

Nice problem.

-Veit

...

On Jan 30, 2017, at 7:43 PM, Dan Asimov <dasimov@earthlink.net> wrote:

----- QUESTION: Given the square torus T^2, what is the smallest number of triangles in a tiling of the torus by triangles such that they are all acute?

I need to define a few terms for this particular question:

* The "square torus" T^2 is the result of identifying opposite edges of a square. (Equivalently, it is the metric space obtained from the quotient of the plane R^2 by its integer subgroup Z^2.)

* A "tiling of the torus by triangles", for the purposes of this question, is any union of triangles that equals the torus, such that

a) Any two triangles have disjoint interiors,

and AAAA b) No triangle's vertex lies on the interior of any triangle's edge. -----

It's well known that the square can be tiled in this sense with 8 acute triangles and no fewer. The standard example (https://www.ics.uci.edu/~eppstein/junkyard/acute-square/8-square.gif) gives, by identifying opposite edges of the square, a tiling of the square torus using 8 acute triangles.

Note: The most general flat torus is obtained by identifying opposite edges of a parallelogram. Clearly, any non-rectangular flat torus can be tiled in the sense of this question using only 2 acute triangles.

Can the upper bound of 8 for the square torus be improved on? And what about a non-square rectangular torus?

—Dan

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