Hyperbolic tesselations with same edge length?

Higher-dimensional geometry (previously "Polyshapes").

Hyperbolic tesselations with same edge length?

Postby Marek14 » Mon Jan 29, 2018 2:58 pm

If we have two regular hyperbolic tesselations, {p1,q1} and {p2,q2}, under what conditions will they have the same edge length?

It's fairly easy for triangles: if you have an equilateral triangle with inner angle pi/(2*n), you can put 2n of them together to create a 2n-gon with inner angle pi/n; thus, {3,2n} has the same edge length as {2n,n} - {3,8} and {8,4}, {3,10} and {10,5}, {3,12} and {12,6}...

But what are the other solutions? I have recently studied my theory of uniform tesselations again (I should post about it soon, I work on constructing explicit examples of all the tesselations), and to my surprise, I found that {4,8} and {8,6} share the edge length (allowing for tesselations with 4 squares and 3 octagons with unexpected straight lines).
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Re: Hyperbolic tesselations with same edge length?

Postby wendy » Tue Jan 30, 2018 8:40 am

Let e be the edge, a the short-chord and d the diameter of the polygon. Then:

4 cosh²(e(p,q)) = a²(p)d²(q) - 4.

Since d² = 4/(4-a²), we have 4p²/(4-q)² as a function that increases as e does, here p, q = a(p), a(q).

a² converges on 4. d converges on p/2(pi), so the circles representing polygons {p}, {p+1}, {p+2}, ... tend to become even spaced. The squares tend to increase by (p/pi)

Most instances of (p,q) then have an edge that lies between (inf, q-1), and (inf, q), and we call such the main series. Since these all amount to separate points, there are no two main-series tilings with the same edge.

We then map, for each non-main-series figure, an edge marked in terms of the main series. {8,6} for example, is a main series polygon, since we have 3.4142*4 > 4*2.8944 (u, 5). But 4, 8 is not, since 2*6.828 is less than 4*4, and is thus in the main sequence of 6. We then divide the product, here 13.656, by the diameter of the hexagon (4), to get 3.414. We then search the shortchord table for a value of 3.414, and find 8. So {4,8} has the same edge as {8,6}.

An array of a² is then made for values as far as 1800. We then find what non-main-series ones exist, and what occurs against the main-series ones, and also for pairs of non-main-series values that give the same value. We already know from geometry that {3,p} falls with {p,p/2}, so the search starts with {4,p}.

The resulting equalities, are for n<1800

3,6 = 4,4 = 6,3 Euclidean tilings
3,8 = 8,4 This occurs in the lt{4,3,8}, where the octagons of the tC give {8,4} and the triangles {3,8}
3,10 = 5,6 = 10,5 For example, consider {3,3,3,5} where all of these are cross-sections
4,8 = 8,6 This occurs in the laminate based on {4,8,A}, where {4,8} are formed as the edge cross-section, and {8,6} as the diameters of the insc. CO
5,12 = 12,10 No instance recorded in higher dimensions.
3,p = p,p/2.

It is interesting that {8,4} and {4,8} are the only duals to fall in this set.
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the dream we dream together is reality.

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Re: Hyperbolic tesselations with same edge length?

Postby Marek14 » Tue Jan 30, 2018 8:52 am

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