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From: Archimedes Plutonium on 13 Mar 2010 17:59


Archimedes Plutonium wrote:
> I sort of made the waters muddy when I first blurted out this
> conjecture yesterday by using "any shape in 2D". Muddy waters because,
> obviously if that shape is a unit square itself that the unit
> equilateral triangle is not going to tile it. So what I meant to say
> was that given a choice of a unit tiler, because a large number of
> shaped 2D objects will be given to you to tile, the question is what
> "unit tiler in 2D" is the best choice to make? Do we pick a unit
> square? or a unit area circle?
> or a unit right-triangle? This conjecture believes that the best
> choice to make for a unit tiler is the equilateral triangle. For the
> reason that two triangles can be situated to form a parallelogram and
> thus imitating the tiling ability of a square. And if the object is
> roundish, the equilateral triangle is best capable of tiling a rounded
> object since its long pointed end acts as spokes on a wheel, so that a
> circle is better tiled by equilateral triangles than is square tiling.
>
> So let me try to write out the Conjecture as clear and
> logical as possible:
>
> Equilateral Triangle Tiler Conjecture: In 2D, given a large number of
> 2D objects to tile by unit tilers, the unit tiler that achieves the
> maximum tiling is the unit equilateral triangle.
>
> Now I am going to offer some proof of this Conjecture, but before I do
> that I want to remark that in 3D, I wonder if the tetrahedron is the
> maximum unit tiler? I wonder if that is true since, unlike the unit
> sphere that is always going to have gaps and holes between tangent
> spheres, we can pack tetrahedrons as parallelograms in vast reaches of
> 3D as solidly as unit cubes without any gaps in between and then when
> we reach the outer surface of the 3D object with its irregular shapes,
> the pointed ends of the tetrahedron are more likely to fit into those
> irregular shapes.
>
> So I do not know if I can generalize this Conjecture from 2D to 3D
> with equilateral-triangles to tetrahedrons.
> And also, the idea in 3D is that a wedge is the most
> versatile tiling 3D object, for a tetrahedron resembles either a wedge
> or a fulcrum.
>
> And let me touch on the topic of how and why I think this conjecture
> is the equivalent of Least Action Principle in Physics. I believe they
> are equivalent because the residue left over in the packing by
> equilateral triangles is considered to be the residue of
> maximum energy applied to the overall system. So that
> if I can pack the object with maximum unit areas by the equilateral
> triangles, then the residue portion is the reversal of what the "least
> path in physics" is. So the equivalency is the idea that the reverse
> of one is the other.
>
> Possible proofs: of this conjecture of Equilateral Triangle Tiler:
> (A) I already noted that if we are given a fractal geometry of a item
> that is the same throughout as it becomes larger and larger, and if
> the equilateral-triangle unit is the maximum tiler of this fractal
> geometry, implies the proof.
>
> (B) Let me offer an alternative proof scheme. The maximum tiler is
> going to be the maximum tiler of various diameters of circles. So here
> we focus on just
> one object shape but of various diameters. It is an easy proof that
> the hexagon composed of equilateral-triangles is the maximum tiler of
> any given diameter circle. Unit squares or unit circles or unit
> rectangles never come close to matching that 6 unit equilateral-
> triangles tile a circle and all other unit objects will be less than
> 6.
>

Let me offer a possible third alternative proof (C).

We know in Calculus that the area of integration can be viewed as
picket fences summed up
over the graph of the figure. Now picket fences are long slender
rectangles, and in Calculus those rectangles are made as skinny as
possible. And sitting atop those skinny rectangles
forming a picket fence is a triangle. Now noone in math history ever
bothered to focus on those triangles, whether they are equilateral-
triangles or whether they are right-triangles. I have always, in my
mind, assumed they were tiny right-triangles. But what if they were
actually, really better off in being equilateral-triangles.

So here, if equilateral-triangles are better served than right-
triangles for the picket-fences
in the integral of Calculus, would be the fastest and most simple
proof that the equilateral
triangle is the maximum tiler in 2D.

Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
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