Octagons with integral coordinates?

[quickfur]

In 2D, it is impossible to represent the coordinates of an equilateral triangle using only integer coordinates. However, in 3D this is possible, as a face of the alternated cube: (1,1,1), (-1,-1,1), (1,-1,-1), for example.

Similarly, a regular octagon cannot have integer coordinates in 2D; but can it have integer coordinates in 3D or higher? What about regular pentagons?

[wendy]

A regular octagon can not have integer coordinates in any dimension. This is because an chord parallel to an edge would be of length 1+sqrt(2), and the implication here is that sqrt(2) can be expressed in integers.

On the other hand, one can get really close with the sets of points (17,0), and (12,12), with all permutations, change of sign. The area of this figure differs from the octagon by a measure of 1/169 of the unit-square.

[bruce741]

all regular.

[Marek14]

To explain Wendy's statement further: no integer multiple of 1 + sqrt(2) can appear as distance of two points with integer coordinates, regardless of dimension.

Proof:

Let's have one of the points in origin and the other one with coordinates [a1,a2,a3,...an].
The distance is sqrt(a1^2 + a2^2 + a3^2 + ... + an^2).
Assume this distance is a multiple of 1 + sqrt(2), k(1 + sqrt(2)) with k > 0.
We have sqrt(a1^2 + a2^2 + a3^2 + ... + an^2) = k(1 + sqrt(2)).
Since both sides are positive, we can square them:
a1^2 + a2^2 + a3^2 + ... + an^2 = k^2(1 + sqrt(2))^2
When we expand (1 + sqrt(2))^2, we get:

a1^2 + a2^2 + a3^2 + ... + an^2 = k^2(3 + 2 sqrt(2)) = 3 k^2 + 2 k^2 sqrt(2)

By further manipulation, we get

(a1^2 + a2^2 + a3^2 + ... + an^2 - 3 k^2)/(2 k^2) = sqrt(2)

This would express sqrt(2) as a ratio of integers, which is impossible.

[quickfur]

If your argument is correct, then that implies that any ratio of the form a+b*√c where a, b, c ≠ 0 cannot be expressed by two points with integer coordinates in any dimension. In the case of the triangle and hexagon (== truncated triangle), the ratio is of the form b*√(c/d), so this proof doesn't apply.

And indeed, we find that we can express equilateral triangles and hexagons with integer coordinates. And in fact, since the coordinates of any regular n-simplex can be expressed in the form b*√(c/d), given the right orientation, this means that any n-simplex and its Stott expansions/truncations thereof can have integer coordinates in some space of dimension k. And indeed, we find that k=n+1 is sufficient: all n-simplex truncates occur as facets of an (n+1)-cubic truncate, and although initially this seems to negate the proof (since k-cubic truncates can have coordinates of the form a+b*√2), it's actually not a problem because the self-duality of the n-simplex means that half of the (n+1)-cubic truncates are congruent to one of the others, and all unique n-simplex truncates occur as facets of some polytope of the form o4x3y3z3... where the first node in the CD diagram is unmarked, which in turn guarantees that their coordinates have a=0, so they reduce to b*√2, which, upon multiplication by 1/√2, become integers. A particular interesting example of this is the omnitruncated n-simplex, which is none other than the permutahedron of order (n+1). (A permutohedron is a polytope whose coordinates are all permutations of (1,2,3,4,...n), or equivalently, of (0,1,2,3,...(n-1)).)

This then leads to the next question: which of the regular polygons have the property that they can be represented by integer coordinates in a space of some dimension k? So far we have established that the triangle, square, and hexagon have this property, but the octagon does not. Are these 3 the only possibilities, or are there others?

[Marek14]

Well, the necessary (but maybe not sufficient) condition for integer polygon is that ALL distances between its vertices must be expressible as distances of integer points. In practice, an n-gon has n/2 distinct distances, rounded down. Now, any distance between two integer points must be an integer or a square of integer. No other values are permitted. For a triangle, square and hexagon, this is true, but I don't thing it holds for any other polygon.

The octagon argument I showed, I must say, is not completely correct: the k should not be an integer because the edge of the octagon doesn't have to be integer either -- k should be the distance between two adjacent vertices of the octagon. But this doesn't actually matter that much since if k is a distance between two points, we can construct a cuboid from hypercubes of edge k that will have diagonal an integer multiple of (1 + sqrt(2)), so it still holds up in the end.

Let's ask the question: why can we have integer triangles and hexagons in 3D? And one of the answers is that we can take normal cubic lattice and alternate it into tetrahedron/octahedron lattice. That still uses only integer points, and contains whole planes with triangular tilings. We can also freely rectify tilings -- but not truncate, as that can introduce noninteger points. Except that triangles can be truncated into hexagons because that divides the sides of triangle in 3 equal parts.

So you can see why not pentagons, heptagons or anything like that: there is no regular lattice, in any dimension, that would contain such shapes in the first place. Octagons and dodecagons can be probably eliminated by solving equations for truncating squares and hexagons.

Now the questions are what integer polytopes can be built from these. Prisms are of course doable. I suspect that all polytopes from simplex families are possible, and all polytopes from cube/orthoplex families that don't include octagons (and are not snub). Since demicubes and Gosset polytopes have no 4 branches, they should all exist in integer forms as well.

Moreover, since n-cube/orthoplex can be made integral in n-dimensions and n-simplex in n+1 dimensions, the same should hold for everything derived from them.

I suspect that square or hexagonal antiprism can't be made integer.

[quickfur]

I think your argument about lattices is not quite waterproof. There *are* space tessellations involving the pentagonal polytopes, for example. They aren't regular, of course, but they do exist. So the non-existence of a suitable lattice is not a necessarily what makes something impossible.

[quickfur]

It's interesting that the demicubes and Gosset polytopes can be integral... can their respective truncates be integral too? Or only a subset of them?

[Marek14]

The pentagonal polytopes aren't impossible because there would be no possible tiling with them, but rather because this tiling can't be derived from the basic cubic tiling. There is a dearth of regular tilings; in 2D there are triangular and hexagonal tilings which can only have integral coordinates when the plane is put into 3D space and in 4D you have 16-cell and 24-cell tiling which can be both derived from tesseractic tiling. (16-cell tiling is alternated tesseractic and 24-cell tiling is the dual of 16-cell.) Basically, my argument was why triangle/hexagon can be integer, not why the other can't.

In 6 to 8 dimensions you have tilings based on Gosset polytopes.

I realized that the set of cubic/orthoplex polytopes with integer coordinates is smaller than I thought, btw. For example, you can't have integer rhombicuboctahedron because it has octagons hidden inside.

So in 3D, you have:

Cube - (1,1,1) with sign changes; (1,1,1) joined to (-1,1,1), (1,-1,1) and (1,1,-1)
Octahedron - (1,0,0) with sign changes and permutations; (1,0,0) joined to (0,1,0), (0,-1,0), (0,0,1) and (0,0,-1)
Cuboctahedron - (1,1,0) with sign changes and permutations; (1,1,0) joined to (1,0,1), (1,0,-1), (0,1,1) and (0,1,-1)
Truncated octahedron - (2,1,0) with sign changes and permutations; joined to (2,0,1), (2,0,-1) and (1,2,0)

In 4D, you have:
1000 - tesseract - (1,1,1,1) with sign changes; (1,1,1,1) joined to (-1,1,1,1), (1,-1,1,1), (1,1,-1,1) and (1,1,1,-1)
0100 - rectified tesseract - (1,1,1,0) with permutations and sign changes; (1,1,1,0) joined to (1,1,0,1), (1,1,0,-1), (1,0,1,1), (1,0,1,-1), (0,1,1,1) and (0,1,1,-1)
0010 - 24-cell - (1,1,0,0) with permutations and sign changes; (1,1,0,0) joined to (1,0,1,0), (1,0,-1,0), (1,0,0,1), (1,0,0,-1), (0,1,1,0), (0,1,-1,0), (0,1,0,1) and (0,1,0,-1)
0001 - 16-cell - (1,0,0,0) with permutations and sign changes; (1,0,0,0) joined to (0,1,0,0), (0,-1,0,0), (0,0,1,0), (0,0,-1,0), (0,0,0,1) and (0,0,0,-1)
0110 - bitruncated tesseract - (2,2,1,0) with permutations and sign changes; (2,2,1,0) joined to (2,2,0,1), (2,2,0,-1), (2,1,2,0) and (1,2,2,0); note that this contains truncated tetrahedra
0101 - rectified 24-cell - (2,1,1,0) with permutations and sign changes; (2,1,1,0) joined to (2,1,0,1), (2,1,0,-1), (2,0,1,1), (2,0,1,-1), (1,2,1,0) and (1,1,2,0)
0011 - truncated 16-cell - (2,1,0,0) with permutations and sign changes; (2,1,0,0) joined to (2,0,0,1), (2,0,0,-1), (2,0,1,0), (2,0,-1,0) and (1,2,0,0)
0111 - truncated 24-cell - (3,2,1,0) with permutations and sign changes; (3,2,1,0) joined to (3,2,0,1), (3,2,0,-1), (3,1,2,0) and (2,3,1,0)

[quickfur]

[...]I realized that the set of cubic/orthoplex polytopes with integer coordinates is smaller than I thought, btw. For example, you can't have integer rhombicuboctahedron because it has octagons hidden inside.[...]

Of course. I don't know if you recall the uniform polytope coordinates derivation scheme from CD diagrams that I posted some time ago. Basically, given any CD diagram of an n-cubic uniform polytope, I can produce coordinates for them instantly just by reading the CD diagram. Given a diagram .4.3.3..., the single node at the end of the edge marked 4 contributes 1 to the resulting coordinates if it is marked, and every subsequent node contributes √2 to the previous accumulated value if it is marked, otherwise it's zero. So x4o3o, for example, would generate the coordinates (1,1,1) (because x4 contributes 1, then 4o3 adds 0, so the 2nd coordinate simply copies the first, and 3o also adds 0, so the 3rd coordinate also just copies the second). SImilarly, o4x3o produces (0, √2, √2), and o4x3x produces (0, √2, 2√2), etc..

If you consider what kind of coordinates are produced, you can immediately tell that anything that starts with o4... will have coordinates of the form k√2, which, since √2 becomes a common factor, you just divide away and you get integer coordinates. On the other hand, anything that starts with x4 produces coordinates of the form 1+k√2, so the only case that's going to give you integer coordinates is when k=0, which gives you only the n-cubes. Anything that has a ringed node after the x4 will give you 1+k√2 with non-zero k, of which the √2 factor cannot be eliminated, so it's impossible to get integer coordinates from it.

So this tells us that the only n-cube family uniform polytopes that can have integer coordinates are of the form o4.3.3...3. or x43o3o3o3o...3o.

Furthermore, by deleting the .4 node from the diagram, you get an (n-1)-simplex family uniform truncate. By extension, then, any n-cube truncate that has integer coordinates will also produce an (n-1)-simplex truncate with integer coordinates. Since the o4.3.3.... form of n-cube truncates allow arbitrary x or o marking of all nodes after the o4 node, this means all (n-1)-simplex truncates (including the simplex itself) can have integer coordinates. Therefore, we conclude that all uniform simplex truncates can have integer coordinates (albeit in (n+1)-space; in their "native" n-space they require radicals of the form √(a/b) with the values of a and b ranging over the square/triangular numbers, respectively, so it is not possible to eliminate them all except for n=1.).

[quickfur]

Now, it's interesting that you mention the Gosset polytopes; it makes me wonder if it's possible to derive larger regular polygons from them? Also, what subset of their truncates continue to allow integral coordinates?

A further line of thought proceeds as follows: one could argue that the reason for the apparent arbitrary preference for n-cubes and the incidental n-simplices (or, if you like, squares and triangles), is that our chosen coordinate system is unduly biased in their preference. So the next question is, suppose, instead of using regular Cartesian coordinates, we use a "skewed" coordinate system instead, say in 3D we choose 3 edges from the tetrahedron that share the same vertex and use them to establish the basis vectors of our coordinate system. Then, obviously, the tetrahedron will have integral coordinates in 3D (we wouldn't need to go to 4D to get this), under this new coordinate system. But would this allow us to also express all the other tetrahedron truncates with integer coordinates? Would the cubic truncates still have integer coordinates under this system?

Is there a choice of basis vectors that would give us integral coordinates for the pentagonal polytopes? Can all of them be expressed using a minimal basis set (i.e., only 3 vectors allowed in 3D)?

Is there a choice of (minimal) basis vectors that would give us integral coordinates for octagonal polytopes?

If so, what is the (minimal) basis set that would give us the maximal number of regular polygons with integer coordinates?

[wendy]

The ''eutactic star" of a mirror-group is formed by unit vectors perpendicular to each of the mirror planes. The eutactic lattice is then formed by placing a parallel eutactic star at the ends of any open tips.

The interesting thing is that any wythoff mirror-edge polytope of integer edge, has its vertices on the eutactic lattice, and therefore must have chord distances which are a subset of the eutactic lattice. We can then see that if the nodes of a wythoff symbol are marked with integer branches, the distances between any two vertices must also be the square root of an integer. The only regular polygons that pass this test are Wn, where n is an integer (ie triangle, square, hexagon, horogon, W5, etc).

When there is a 4-branch or a 6-branch in the picture, you do get A+Bq or A+Bh, where q²=2, h²=3. But if all the nodes on one side of the 4-branch are marked q, then the vertices become integer again. So o4o3q and x4o3o fall on the same lattice. The Lie group people mark a four branch as q==>==x (double bar and arrow, pointing from the q to the x), since they don't admit integers from Z4 (A+Bq) or Z6 (A+Bh)

The 5-branch creates an integer system in the pentagonal numbers Z5 = A+Bf.

The tiling of octagons, or of octagrammy in 4D, produce a cover in Z4 (the span of chords of the square, or A+Bq), while the dodecagonal tiling is in Z6 (hexagon-chords, A+Bh). But all of the A+Bx, of Z4, Z5, Z6, are class 2 (infinitely dense as the decimals are, that is, mapping a 2d lattice onto the number line).