Yep, you nailed it. I'm guessing Allan did, too, since his 4-digit figure of 3.8637+ agrees with what I got (√2 + √6 = (3.863703305156273+) for the boundary length, using the same non-regular hexagon you mention. However, I don't have a proof handy just now. —Dan William Somsky wrote: ----- Hexagons (not rotationally symmetric) with 120 degree interior angles and a-c = 1 unit, b-c = 1 unit, a-c perpendicular b-c a___ / \ / \c \ / \b____/ On Sun, Aug 9, 2020, 15:30 Allan Wechsler <acwacw@gmail.com> wrote:
A minimum must exist, very roughly, because each tile has unit area, and the minimum circumference for a thingy of unit area is 2 * sqrt(pi), which is 3.5449+.
The circumference achieved by a square tiling is 4. I have a trick to get as low as 3.8637+, and I have a strong feeling that it's the best trick, but am nowhere near a proof.
On Sun, Aug 9, 2020 at 6:03 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Nit-picking:
"each tile to a tile" perhaps?
Unclear what deformations are implied by "adjust"!
Finally, (why) must such a minimum exist?
WFL
On 8/9/20, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose we tessellate the plane by the unit squares with vertices at the integer points. We'd now like to adjust this tessellation so that
1) The boundary of each tile is a simple closed polygon;
2) The set of translations that carry each tile to another tile is exactly the integer vectors (K,L) ∊ Z^2.
Puzzle: Which tile shape satisfying 1) and 2) is such that each tile's boundary polygon has the shortest possible length?
—Dan