What I meant was: take your torus, cut and unroll it to be your square ABCD. Now, as a square, clip off BCE and glue it to the other side with BC on AD to make a parallelogram. Divide it on its diagonal as roll it back into a torus. But now, spelling it out, I see my flaw: identifying the edges of the new parallelogram, CD wouldn't get pasted back onto BA, but into the shifted EABE. Yeah, I think it's the same as you spelled out or maybe a "dual" where I don't violate the triangle rule but end up creating a different torus. On Jan 30, 2017 21:37, "Dan Asimov" <dasimov@earthlink.net> wrote: WRS, I don't know if I'm understanding this from your description. I'm thinking of the square being say ABCD (ccw starting from upper right). B————E————A | / \ | | / \ | | / \ | |/ \| C—————————D Then say let E be the midpoint of AB and F be the midpoint of CD. Finally one triangle is CED and the other is what remains: the triangles EBC and AED joined along their common edge BC = AD by the identification of the square to make a torus. IF that is the same idea as you are suggesting, it has the flaw that the vertex E (of triangle CED) is identified with an interior point of the edge CD. Which violates condition b) below. Or am I misunderstanding? —Dan
On Jan 30, 2017, at 8:13 PM, William R Somsky <wrsomsky@gmail.com> wrote:
Two, I believe. Unless I'm mistaken, you just trim a triangle off the left edge of the square, move it to the right side matching the identified edges, and you have a parallelogram, which is trivially two-acute-triangularizable.
On Jan 30, 2017 16:44, "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
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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