User:Harry Princeton/Deltoidal Icositetrahedron

The deltoidal icositetrahedron is the Catharry solid with kite faces, and with cube symmetric group.

Properties (Comparison)
In the construction section below, we shall formally derive these. and the Center of the icositetrahedral deltoid is 53.69% the distance between corners Vertex I and Vertex III, or the light black vertex between the red and yellow vertices as below:



Construction
It can be made from the polar reciprocation of the recticuboctahedron, the Archimedean solid, about its midsphere (the rectified cuboctahedron with rectangles lining the faces homeomorphic to the cuboctahedron). If the recticuboctahedron has side length 1, then$$r_{\text{mid}}=\sqrt{1+\frac{1}{\sqrt{2}}}\approx 1.3066=R_{\text{mid}}~r_{\text{cir}}=\frac{\sqrt{5+2\sqrt{2}}}{2}\approx 1.3990~R_{\text{in}}=\sqrt{\frac{14+8\sqrt{2}}{17}}\approx 1.2203$$

where $$r_{\text{mid}}=R_{\text{mid}}$$ is the common midradius, $$r_{\text{cir}}$$ is the circumradius of the recticuboctahedron, and $$R_{\text{in}}$$ is the inradius of the deltoidal icositetrahedron.

All of this can be derived from the equatorial regular octagon of the same side length, from the recticuboctahedron (this is really a girth; the deltoidal icositetrahedron's regular octagon is actually an equator with side length $$\sec\frac{\pi}{8}=\sqrt{2\left(2-\sqrt{2}\right)}\approx 1.0824$$).

Moreover, the square and triangular apothems of the recticuboctahedron are

$$\text{apo}_{\Box}=\frac{1+\sqrt{2}}{2}\approx 1.2071~\text{apo}_{\triangle}=\frac{3+\sqrt{2}}{2\sqrt{3}}\approx 1.2743$$

Polar reciprocation (actual geometric inversion) of these apothems yield the corresponding deltoidal icositetrahedron's perithems of (as in the aphelion and perihelion, and perimeter):

$$\text{PERI}_{\Box}=\sqrt{2}\approx 1.4142~\text{PERI}_{\triangle}=\frac{4\sqrt{3}+\sqrt{6}}{7}\approx 1.3397$$

Conversely, $$r_{\text{cir}}\approx 1.3990$$ is the unique perithem of the recticuboctahedron (rco), and $$R_{\text{in}}\approx 1.2203$$ is the unique apothem of the corresponding deltoidal icositetrahedron (DI).

Taking the deltoid in the first octant closest to the $$x$$-axis, we have the Euclidean vertices:

$$\text{red}:\left(\sqrt{2},0,0\right)~\text{dark red}:\left(1,1,0\right)~\text{yellow}:\left(\frac{4+\sqrt{2}}{7},\frac{4+\sqrt{2}}{7},\frac{4+\sqrt{2}}{7}\right)~\text{dark red}:\left(1,0,1\right)$$

given the DI is polar-reciprocated from the rco of side length 1 (counterclockwise from top of $$x$$-axis). Then the deltoid's apothem located at

$$\left(\frac{8+7\sqrt{2}}{17},\frac{6+\sqrt{2}}{17},\frac{6+7\sqrt{2}}{17}\right)$$

Numerically, the five points on the deltoid are approximately $$(1.4142,0,0)-(1,1,0)-(0.7735,0.7735,0.7735)-(1,0,1)$$ and $$(1.0529,0.4361,0.4361)$$.

Passing the deltoid into the Cartesian plane and using the Pythagorean Theorem, the diagonals' side lengths are computed as


 * $$\text{diag}_{\text{dark red},\text{dark red}}=\sqrt{2}\approx 1.4142~\text{diag}_{\text{red},\text{yellow}}=\frac{2\sqrt{31-8\sqrt{2}}}{7}\approx 1.2677$$

which implies the deltoid is wider than it is high; compare the deltoids in the deltoidal trihexagonal tiling, which are higher than wide ($$1+\frac{1}{\sqrt{3}}\approx 1.5774$$ by $$\frac{1+\sqrt{3}}{2}\approx 1.3660$$): Moreover, by passing the icositetrahedral's deltoid's perithem to the same plane, we see that the apothem point lay (which is an incenter, because deltoids are cocyclic!):


 * $$\frac{11-\sqrt{2}}{17}\approx 0.5369$$

of the distance from the red to the yellow point, by some computations (compare to the trihexagonal deltoid's incenter at $$\frac{3-\sqrt{3}}{2}\approx 0.6340$$ of the distance).

Furthermore, the two diagonals' intersection lay


 * $$1-\frac{1}{2\sqrt{2}}\approx 0.6464$$

from the red to the yellow point (versus trihexagonal: $$\frac{3}{4}=0.7500$$).

Finally, we compute the side lengths of the icositetrahedral deltoid (and thus the DI) as


 * $$\Box\text{-}\Box=\sqrt{4-2\sqrt{2}}\approx 1.0824=\sec\frac{\pi}{8}~\Box\text{-}\triangle=\frac{2\sqrt{10-\sqrt{2}}}{7}\approx 0.8372$$

and the ratio of the side lengths is


 * $$\frac{\Box\text{-}\Box}{\Box\text{-}\triangle}=\frac{7\sqrt{4-2\sqrt{2}}}{2\sqrt{10-\sqrt{2}}}=\frac{7}{2}\sqrt{\frac{4-2\sqrt{2}}{10-\sqrt{2}}}=\sqrt{\frac{\left(4-2\sqrt{2}\right)\left(10+\sqrt{2}\right)}{8}}=\sqrt{\frac{9-4\sqrt{2}}{2}}=\sqrt{\frac{\left(2\sqrt{2}-1\right)^2}{2}}=2-\frac{1}{\sqrt{2}}\approx 1.2929$$

as computed before (versus trihexagonal, sides ratio $$\frac{1+\sqrt{3}}{2}\Bigg/\frac{1+\sqrt{3}}{2\sqrt{3}}=\sqrt{3}\approx 1.7321$$).

Spherical Angles


The geodesic arc angles of the spherical deltoid sides are


 * $$\Box\text{-}\Box=45^{\circ}~\Box\text{-}\triangle=\sin^{-1}\frac{1}{\sqrt{3}}=35.26^{\circ}$$

with the first case determined by the equatorial regular octagon, and the second case as an angle of the line $$x=y=z$$ (one deltoid perithem - yellow point) with $$x=y,z=0$$.

The geodesic arc angles of the spherical deltoid diagonals are


 * $$\text{diag}_{\text{dark red},\text{dark red}}= 60^{\circ}\text{diag}_{\text{red},\text{yellow}}=\sin^{-1}\sqrt{\frac{2}{3}}\approx 54.74^{\circ}$$

with the first case occuring since the dark red points are exactly $$\sqrt{2}$$ apart, and the Euclidean norm of those points is $$\sqrt{2}$$ (there is an equilateral triangle between dark red points and center); and the second case determined as the angle of the deltoid apothem with the line $$x=y,z=0$$.

The spherical apothem is


 * $$\frac{\sin^{-1}\sqrt{\frac{10-4\sqrt{2}}{17}}}{\sin^{-1}\sqrt{\frac{2}{3}}}\approx 0.5547$$

of the way from the red to the yellow point, on the main diagonal.

By intersecting the planes $$y=z$$ (containing the spherical dark red points and the center) and $$x-y-z=0$$ (containing the spherical red and yellow points, and the center); with the sphere, we see that the geodesic diagonals intersect at a point


 * $$\frac{\sin^{-1}\frac{1}{\sqrt{3}}}{\sin^{-1}\sqrt{\frac{2}{3}}}\approx 0.6443$$

of the way from the red to the yellow point, on the main diagonal.

Cocyclic Generalization - Catalan (Uniform) Spherical Circle Packings
Note that the faces of the deltoidal icositetrahedron are cocyclic; so are the faces of the deltoidal hexecontahedron, rhombic dodecahedron, and rhombic triacontahedron (since rhombi are kites). Any triangle is cocyclic, so the faces of the triakis tetrahedron, triakis octahedron, tetrakis cube, triakis icosahedron, pentakis dodecahedron, disdyakis dodecahedron, and disdyakis triacontahedron are cocyclic.

It remains to show whether the faces of the pentagonal icositetrahedron and pentagonal hexecontahedron (Cairo/Floret solids) are, in fact, cocyclic. If so, we can homeomorphically project those Catalan solids onto spheres and realize their Spherical Circle Packings as well. If this is the case, we can size up every Catalan solid by setting corresponding spherical circles (which are actual circles) at the same radius, preferrably $$1/2$$ from ambo circle packings of the Euclidean plane; along with comparing them using Archimedean solids of the same side length and polar-reciprocating them.

Measures of the faces show the each pentagon has one axis of symmetry, based on the configuration vertices: Meanwhile, realizing the intersections of the Catalan edges with the Archimedean edges generates cyclic spherical hCVRPs (technically hCVRSs: half-size configuration vertex regular spherigons), which implies the cocyclicity of the Catalan pentagons (homeomorphic to vertex regular spherigons).

Furthermore, the Catalan and Archimedean edges of dual compounds are perpendicular, and perpendicularity is invariant under dihedral angle homeomorphisms; over each cyclic hCVRS is a shallow pyramid (with hCVRS base), which can be circumscribed by a shallow cone; this cone yields the circle for the spherical circle packing if the circular base is projected to the midsphere.

Hence, all Catalan spherical circle packings can be realized; see below for a circle packing (blue on white) on the polyhedron itself (before homeomorphism) - not a Johnson Solid Circle Packing since the blue circles do not have equal angular area:

Result
This means, we can recover the original uniform tiling from the dual uniform tiling by connecting incenters of planigons, spherigons, or hyperboligons, among shared edges; since they are all cocyclic.

Conversely, constructing a dual uniform tiling is easy, as all regular (convex) polygons have centroids which coincide with incenters, orthocenters, circumcenters, etc.

See below for polyhedral spherigons and spherigonal circle packings.

Catalaves Polygons (Polyhedral Spherigons)
For each spherigon that can tile the sphere alone, there is a corresponding Catalaves polygon (the polygonal face of a Catalan Solid), or a polyhedral vertex regular spherigon (PVRS). There are 3 regular Catalaves polygons (e.g. regular Platonic polygons) and 13 semiregular Catalaves polygons (more than the number of semiregular planigons!). They are listed below:

Regular and Uniform Spherigonal Circle Packings
There are 5 regular spherigonal circle packings (corresponding to the Platonic solids), and 13 semiregular (uniform) spherigonal circle packings (corresponding to the 13 Catalan solids); tabulated as above (and with densities below). The circles are constructed by the projection of incircles of the cocyclic Catalaves tiles from the Catalan solid up to the midsphere (thus inscribing circles in the vertex-regular spherigons themselves, on the midsphere; these are circumcircles of half-size configuration vertex Catalaves tiles). The densest uniform spherigonal circle packing is the dodecahedral one at 89.60951%; the lightest uniform spherigonal circle packing is the triakis icosahedral one at 42.83424%.

Compare these extremes to the uniform circle packings of the Euclidean plane:

$$\rho_{\text{hexagonal}}=\frac{\pi}{2\sqrt{3}}\approx 90.68997\%~\rho_{\text{kisrhombille}}=\pi\left(7\sqrt{3}-12\right)\approx 39.06748\%$$

To compute the density of these spherigonal circle packings, take the ratio of the solid angle occupied by all circles to the whole sphere of area $$4\pi$$ steradians: (this kind of triple right-triangle treatment was used in constructing ambo circle packings in the 2-D case). From the diagrams above, we see that the radius of the circle inscribed in the Catalaves tile (or circumscribing the half-size configuration Catalaves polygon) is


 * $$r^*=\frac{r_{\text{mid}}}{2r_{\text{cir}}}$$

And the (half) apex angle of the cone is


 * $$\theta^*=\sin^{-1}\frac{1}{2r_{\text{cir}}}$$

Using the apex area formula $$s=4\pi\sin^2\frac{\theta}{2}$$, we have


 * $$s^*=4\pi\sin^2\frac{\theta^*}{2}=4\pi\left(\frac{1-\cos\theta^*}{2}\right)=4\pi\left(\frac{1-\sqrt{1-\frac{1}{4r_{\text{cir}}^2}}}{2}\right)=4\pi\left(\frac{1}{2}-\frac{\sqrt{4r_{\text{cir}}^2-1}}{4r_{\text{cir}}}\right)=\frac{1}{2}-\frac{\sqrt{4r_{\text{cir}}^2-1}}{4r_{\text{cir}}}~\text{of the sphere}$$

by the trigonometric identity of $$\sin(\theta/2)$$. In fact, such a circle carves out a spherical dome of that solid angle area precisely on the Archimedean solid's midsphere. Thus, if the corresponding Archimedean solid has $$V$$ vertices and circumradius $$r_{\text{cir}}$$ (and the Catalan solid with $$V$$ faces), the solid-angle density of the spherigonal circle packing is


 * $$V\left(\frac{1}{2}-\frac{\sqrt{4r_{\text{cir}}^2-1}}{4r_{\text{cir}}}\right)=V\left(\frac{1}{2}-\frac{r_{\text{mid}}}{2r_{\text{cir}}}\right)$$

since $$r_{\text{mid}}^2+(1/2)^2=r_{\text{cir}}^2$$(the edges are tangent to the midsphere of an Archimedean solid).

Using this formula, we compute the solid-angle densities of the regular (5) and semiregular (13) spherigonal circle packings.

Densities of 5 Regular Spherigonal Circle Packings


 * 1) $$\bullet:~\rho_{\text{Tetrahedron}}=4\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{6}}{4}\right)^2-1}}{4\left(\frac{\sqrt{6}}{4}\right)}\right)=2-\frac{2}{\sqrt{3}}\approx 0.8452995$$
 * 2) $$\bullet:~\rho_{\text{Octahedron}}=8\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{3}}{2}\right)^2-1}}{4\left(\frac{\sqrt{3}}{2}\right)}\right)=4-4\sqrt{\frac{2}{3}}\approx 0.7340137$$
 * 3) $$\bullet:~\rho_{\text{Cube}}=6\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{1}{\sqrt{2}}\right)^2-1}}{4\left(\frac{1}{\sqrt{2}}\right)}\right)=3-\frac{3}{\sqrt{2}}\approx 0.8786797$$
 * 4) $$\bullet:~\rho_{\text{Icosahedron}}=20\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{3}+\sqrt{15}}{4}\right)^2-1}}{4\left(\frac{\sqrt{3}+\sqrt{15}}{4}\right)}\right)=10-10\sqrt{\frac{3+\sqrt{5}}{6}}\approx 0.6582764$$
 * 5) $$\bullet:~\rho_{\text{Dodecahedron}}=12\left(\frac{1}{2}-\frac{\sqrt{4\left(\sqrt{\frac{5+\sqrt{5}}{8}}\right)^2-1}}{4\left(\sqrt{\frac{5+\sqrt{5}}{8}}\right)}\right)=6-3\sqrt{2+\frac{2}{\sqrt{5}}}\approx 0.8960951$$

Densities of 13 Uniform Spherigonal Circle Packings


 * 1) $$\bullet:~\rho_{\text{Triakis Tetrahedron}}=12\left(\frac{1}{2}-\frac{\sqrt{4\left(\sqrt{\frac{11}{8}}\right)^2-1}}{4\left(\sqrt{\frac{11}{8}}\right)}\right)=6-\frac{18}{\sqrt{11}}\approx 0.5727958$$
 * 2) $$\bullet:~\rho_{\text{Triakis Octahedron}}=24\left(\frac{1}{2}-\frac{\sqrt{4\left(\sqrt{\frac{7}{4}+\sqrt{2}}\right)^2-1}}{4\left(\sqrt{\frac{7}{4}+\sqrt{2}}\right)}\right)=12-12\sqrt{\frac{10+4\sqrt{2}}{17}}\approx 0.4838042$$
 * 3) $$\bullet:~\rho_{\text{Tetrakis Cube}}=24\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{10}}{2}\right)^2-1}}{4\left(\frac{\sqrt{10}}{2}\right)}\right)=12-18\sqrt{\frac{2}{5}}\approx 0.6158004$$
 * 4) $$\bullet:~\rho_{\text{Triakis Icosahedron}}=60\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{74+30\sqrt{5}}}{4}\right)^2-1}}{4\left(\frac{\sqrt{74+30\sqrt{5}}}{4}\right)}\right)=30-30\sqrt{\frac{85+15\sqrt{5}}{122}}\approx 0.4283424$$
 * 5) $$\bullet:~\rho_{\text{Pentakis Dodecahedron}}=60\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{58+18\sqrt{5}}}{4}\right)^2-1}}{4\left(\frac{\sqrt{58+18\sqrt{5}}}{4}\right)}\right)=30-90\sqrt{\frac{21+\sqrt{5}}{218}}\approx 0.6170374$$
 * 6) $$\bullet:~\rho_{\text{Rhombic Dodecahedron}}=12\left(\frac{1}{2}-\frac{\sqrt{4\left(1\right)^2-1}}{4\left(1\right)}\right)=6-3\sqrt{3}\approx 0.8038476$$
 * 7) $$\bullet:~\rho_{\text{Rhombic Triacontahedron}}=30\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{1+\sqrt{5}}{2}\right)^2-1}}{4\left(\frac{1+\sqrt{5}}{2}\right)}\right)=15-\frac{15}{2}\sqrt{\frac{5+\sqrt{5}}{2}}\approx 0.7341523$$
 * 8) $$\bullet:~\rho_{\text{Deltoidal Icositetrahedron}}=24\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{5+2\sqrt{2}}}{2}\right)^2-1}}{4\left(\frac{\sqrt{5+2\sqrt{2}}}{2}\right)}\right)=12-12\sqrt{\frac{12+2\sqrt{2}}{17}}\approx 0.7926140$$
 * 9) $$\bullet:~\rho_{\text{Deltoidal Hexecontahedron}}=60\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{11+4\sqrt{5}}}{2}\right)^2-1}}{4\left(\frac{\sqrt{11+4\sqrt{5}}}{2}\right)}\right)=30-30\sqrt{\frac{30+4\sqrt{5}}{41}}\approx 0.7617671$$
 * 10) $$\bullet:~\rho_{\text{Disdyakis Dodecahedron}}=48\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{13+6\sqrt{2}}}{2}\right)^2-1}}{4\left(\frac{\sqrt{13+6\sqrt{2}}}{2}\right)}\right)=24-24\sqrt{\frac{84+6\sqrt{2}}{97}}\approx 0.5651766$$
 * 11) $$\bullet:~\rho_{\text{Disdyakis Tricontahedron}}=120\left(\frac{1}{2}-\frac{\sqrt{4\left(\frac{\sqrt{31+12\sqrt{5}}}{2}\right)^2-1}}{4\left(\frac{\sqrt{31+12\sqrt{5}}}{2}\right)}\right)=60-60\sqrt{\frac{210+12\sqrt{5}}{241}}\approx 0.5209986$$
 * 12) $$\bullet:~\rho_{\text{Pentagonal Icositetrahedron}}=24\left(\frac{1}{2}-\frac{\sqrt{4\left(\sqrt{\frac{3-T}{4\left(2-T\right)}}\right)^2-1}}{4\left(\sqrt{\frac{3-T}{4\left(2-T\right)}}\right)}\right)=12-\frac{12\sqrt{3}}{\sqrt{8-\sqrt[3]{19-3\sqrt{33}}-\sqrt[3]{19+3\sqrt{33}}}}\approx 0.8617035$$
 * 13) $$\bullet:~\rho_{\text{Pentagonal Hexecontahedron}}=60\left(\frac{1}{2}-\frac{\sqrt{4\left(2.1558373751156397\right)^2-1}}{4\left(2.1558373751156397\right)}\right)=30-15\sqrt{4-\frac{1}{\left(2.1558373751156397\right)^2}}\approx 0.8180145$$

Vertex Regular Johnson Polygons (VRJPs)
If polar reciprocation of the Johnson solids are possible, then we have vertex regular Johnson Polygons (VRJP, or VRJS for spherigonal version if the Johnson solid has a midsphere) forming the duals of Johnson solids.

One drawback of polar reciprocation (versus topological duality) is that apothems may coincide outside the Johnson solid, or conversely, apothems are beyond the faces given a center point. Apothems are defined for every center (which is the distance from the center point to the plane of the face).

Notably, if the Johnson solid does not have a midsphere, we may not be able to explicitly recover the Johnson solid from its dual, since the polyhedral vertex regular spherigon (PVRS) may not be cocyclic.

There are 92 Johnson solids, and probably around 100 (many non-cocyclic) VRJPs, e.g. V32.5 for the pentagonal pyramid is one of them.

Johnson (k-Uniform) Spherical Circle Packings
If the polar reciprocal of a Johnson solid has all cocyclic polygons, and the Johnson solid is itself cyclic, then we can make the corresponding Johnson Spherical Circle Packing. Though this is quite rare; we can do this for the gyrate rhombicosidodecahedron, for example.

Hyperbolic Plane Circle Packings
Likewise, uniform and k-uniform circle packings of the hyperbolic plane exist aplenty (corresponding to Euclidean tilings), as the Möbius transformations governing the hyperbolic plane map circles onto circles (Poincare model).

Terminology of Tilings
We have (convex) vertex-regular:


 * Planigons (or Laves tilings) for dual uniform planar tilings
 * Spherigons for dual uniform spherical tilings
 * Hyperboligons for dual uniform hyperbolic-plane tilings
 * Stereohedrons for dual uniform honeycombs
 * Stereochorons for dual uniform 4-cell fillings of $$\mathbb{R}^4$$
 * Stereotopes for dual uniform k-cell fillings of $$\mathbb{R}^k$$

Coincidence Between Duality and Polar Reciprocation
See User:Harry Princeton/Planigons and Dual Uniform Tilings for main result.

For the Euclidean plane, reciprocating the vertex-regular planigons (VRPs) of dual uniform tilings about circles of radius $$1/2$$ centered at the vertices; yield half-size configuration vertex-regular polygons (hCVRPs). These hCVRPs tile the plane in checkerboard fashion, with rectified regular polygons as gap checkers. The result is an ambo uniform tiling (or rectified uniform tiling). Hence, we can illustrate duality as polar reciprocation by using circle ambo packings with faint dual uniform lattices.

This also works for spherical tilings (and even such polyhedra; see Dorman-Luke construction), and for hyperbolic tilings as well.

However, when we get to dimension 3 or higher, rectification does not yield a topological dual - but an ambo polyhedron (or ambo polychoron or ambo polytope). So even though there are analogous ambo spherical packings for sterohedrons (i.e. like the rhombic dodecahedron for the densest such packing), such packings do not illustrate duality as polar reciprocity (i.e. we get recticuboctahedra, not cuboctahedra).

Conversely, polar-reciprocating sterohedrons about their inspheres may not give rise directly to actual honeycombs, and new polyhedrons must be artificially added (e.g. cuboctahedra only go with octahedra in a cubic lattice, not a rhombic dodecahedral lattice; on the other hand, cuboctahedra are not polar reciprocals of cubes, although they do fill gaps between octahedra with new splitting faces).