pentagonal hexecontahedron

File:Pentagonal hexecontahedron.stl

The faces are irregular pentagons with two long edges and three short edges. Let $\backslash xi\backslash approx\; 0.943\backslash ,151\backslash ,259\backslash ,24$ be the real zero of the polynomial $x^3+2x^2-\backslash phi^2$. Then the ratio $l$ of the edge lengths is given by:$$l\; =\; \backslash frac\{1+\backslash xi\}\{2-\backslash xi^2\}\backslash approx\; 1.749\backslash ,852\backslash ,566\backslash ,74.$$The faces have four equal obtuse angles and one acute angle (between the two long edges). The obtuse angles equal $\backslash arccos(-\backslash xi/2)\backslash approx\; 118.136\backslash ,622\backslash ,758\backslash ,62^\{\backslash circ\}$, and the acute one equals $\backslash arccos(-\backslash phi^2\backslash xi/2+\backslash phi)\backslash approx\; 67.453\backslash ,508\backslash ,965\backslash ,51^\{\backslash circ\}$. The dihedral angle equals $\backslash arccos(-\backslash xi/(2-\backslash xi))\backslash approx\; 153.2^\{\backslash circ\}$.

Note that the face centers of the snub dodecahedron cannot serve directly as vertices of the pentagonal hexecontahedron: the four triangle centers lie in one plane but the pentagon center does not; it needs to be radially pushed out to make it coplanar with the triangle centers. Consequently, the vertices of the pentagonal hexecontahedron do not all lie on the same sphere and by definition it is not a zonohedron.

To find the volume and surface area of a pentagonal hexecontahedron, denote the shorter side of one of the pentagonal faces as $b$, and set a constant{{Cite web|title=Pentagonal Hexecontahedron - Geometry Calculator|url=https://rechneronline.de/pi/pentagonal-hexecontahedron.php|website=rechneronline.de|access-date=2020-05-26}}$$t\; =\; \backslash frac\{\backslash sqrt[3]\{44+12\backslash phi(9+\backslash sqrt\{81\backslash phi-15\})\}+\backslash sqrt[3]\{44+12\backslash phi(9-\backslash sqrt\{81\backslash phi-15\})\}-4\}\{12\}\; \backslash approx\; 0.472.$$Then the surface area ($A$) is:$$A\; =\; \backslash frac\{30b^2\backslash cdot(2+3t)\backslash cdot\backslash sqrt\{1-t^2\}\}\{1-2t^2\}\backslash approx162.698b^2.$$And the volume ($V$) is:$$V\; =\; \backslash frac\{5b^3(1+t)(2+3t)\}\{(1-2t^2)\backslash cdot\backslash sqrt\{1-2t\}\}\backslash approx189.789b^3.$$Using these, one can calculate the measure of sphericity for this shape:$$\backslash Psi\; =\; \backslash frac\{\backslash pi^\{\backslash frac\{1\}\{3\}\}(6V)^\{\backslash frac\{2\}\{3\}\}\}\{A\}\; \backslash approx\; 0.982$$

File:Pentagonal_Hexacontahedron_Convex_Hulls.svg icosahedron (12) centered at the origin with a chiral snub dodecahedron (60) combined with a dodecahedron of the same non-unity circumradius (20) to construct the pentagonal hexecontahedron]]

The pentagonal hexecontahedron can be constructed from a snub dodecahedron without taking the dual. Pentagonal pyramids are added to the 12 pentagonal faces of the snub dodecahedron, and triangular pyramids are added to the 20 triangular faces that do not share an edge with a pentagon. The pyramid heights are adjusted to make them coplanar with the other 60 triangular faces of the snub dodecahedron. The result is the pentagonal hexecontahedron.[https://groups.google.com/g/sci.math/c/c9e37XmKO24/m/okJW1M54kc4J Reference]

An alternate construction method uses quaternions and the icosahedral symmetry of the Weyl group orbits $O(\backslash Lambda)=W(H\_3)/C\_2\; \backslash approx\; A\_5=I$ of order 60.{{cite arXiv

|title=Chiral Polyhedra Derived From Coxeter Diagrams and Quaternions

|first1=Mehmet

|last1=Koca

|first2=Nazife

|last2=Ozdes Koca

|first3=Muna

|last3=Al-Shu’eilic

|year=2011

|class=math-ph

|eprint=1006.3149

}} This is shown in the figure on the right.

Specifically, with quaternions from the binary Icosahedral group $(p,q)\; \backslash in\; I\_h$, where $q=\backslash bar\; p$ is the conjugate of $p$ and $[p,q]:r\backslash rightarrow\; r\text{'}=prq$ and $[p,q]^*:r\backslash rightarrow\; r\text{'}\text{'}=p\backslash bar\; rq$, then just as the Coxeter group $W(H\_4)=\backslash lbrace[p,\backslash bar\; p]\; \backslash oplus\; [p,\backslash bar\; p]^*\backslash rbrace$ is the symmetry group of the 600-cell and the 120-cell of order 14400, we have $W(H\_3)=\backslash lbrace[p,\backslash bar\; p]\; \backslash oplus\; [p,\backslash bar\; p]^*\backslash rbrace=A\_5\backslash times\; C\_2=I\_h$ of order 120. $I$ is defined as the even permutations of $I\_h$ such that $[I,\backslash bar\; I]:r$ gives the 60 twisted chiral snub dodecahedron coordinates, where $r\backslash approx\; -0.389662\; e\_1\; +\; 0.267979\; e\_2\; -0.881108\; e\_3$ is one permutation from the first set of 12 in those listed above. The exact coordinate for $r$ is obtained by taking the solution to $x^3-x^2-x-\backslash phi=0$, with $x\backslash approx\; 1.94315$, and applying it to the normalization of $r=(-1+x^2(-1-2/\backslash phi-x\backslash phi)e\_1\; +\; (3-x^2+3x\backslash phi)e\_2\; +\; ((x^3-1/\backslash phi)\backslash phi^3)e\_3$.

Using the Icosahedral symmetry in the orbits of the Weyl group $O(\backslash Lambda)=W(H\_3)/C\_2\; \backslash approx\; A\_5$of order 60{{cite journal

|title=Catalan Solids Derived From 3D-Root Systems and Quaternions

|first1=Mehmet

|last1=Koca

|first2=Nazife

|last2=Ozdes Koca

|first3=Ramazon

|last3=Koc

|year=2010

|journal=Journal of Mathematical Physics

|volume=51

|issue=4

|doi=10.1063/1.3356985

|arxiv=0908.3272

|s2cid=115157829

}} gives the following Cartesian coordinates with $\backslash phi=\backslash frac\{1+\backslash sqrt\{5\}\}\{2\}$ is the golden ratio:

• Twelve vertices of a regular icosahedron with unit circumradius centered at the origin with the coordinates $$\backslash frac\{(0,\; \backslash pm\; 1,\; \backslash pm\; \backslash phi)\}\{\backslash sqrt\{\backslash phi^2\; +\; 1\}\}\; ,\; \backslash frac\{(\backslash pm\; 1,\; \backslash pm\; \backslash phi,\; 0)\}\{\backslash sqrt\{\backslash phi^2\; +\; 1\}\}\; ,\; \backslash frac\{(\backslash pm\; \backslash phi,\; 0,\; \backslash pm\; 1)\}\{\backslash sqrt\{\backslash phi^2\; +\; 1\}\}.$$
• Twenty vertices of regular dodecahedron of unit circumradius centered at the origin scaled by a factor$R\backslash approx\; 0.95369785218$ from the exact solution to the equation $700569\; -\; 1795770\; x^2\; +\; 1502955\; x^4\; -\; 423900\; x^6\; +\; 14175\; x^8\; -\; 2250\; x^\{10\}\; +\; 125\; x^\{12\}\; =\; 0$, which gives the coordinates

$$(\backslash pm\; 1,\; \backslash pm\; 1,\backslash pm\; 1)\backslash frac\{R\}\{\backslash sqrt\{3\}\}$$ and $$(0,\; \backslash pm\; \backslash phi,\; \backslash pm\; \backslash frac\{1\}\{\backslash phi\})\backslash frac\{R\}\{\backslash sqrt\{3\}\}\; ,\; (\backslash pm\; \backslash frac\{1\}\{\backslash phi\},\; 0\; ,\; \backslash pm\; \backslash phi)\; \backslash frac\{R\}\{\backslash sqrt\{3\}\}\; ,\; (\backslash pm\; \backslash phi,\; \backslash pm\; \backslash frac\{1\}\{\backslash phi\},0)\backslash frac\{R\}\{\backslash sqrt\{3\}\}.$$

• Sixty vertices of a unit circumradius chiral snub dodecahedron scaled by $R$. There are five sets of twelve vertices, all with even permutations (i.e. with a parity signature=1).

A group of two sets of twelve have 0 or 2 minus signs (i.e. 1 or 3 plus signs):

$$(\backslash pm\; 0.267979,\; \backslash pm\; 0.881108,\; \backslash pm\; 0.389662)\; R\; ,$$

$$(\backslash pm\; 0.721510,\; \backslash pm\; 0.600810,\; \backslash pm\; 0.344167)\; R\; ,$$

and another group of three sets of 12 have 0 or 2 plus signs (i.e. 1 or 3 minus signs):$$(\backslash pm\; 0.176956,\; \backslash pm\; 0.824852,\; \backslash pm\; 0.536941)\; R\; ,$$

$$(\backslash pm\; 0.435190,\; \backslash pm0.777765,\; \backslash pm\; 0.453531)\; R\; ,$$

$$(\backslash pm\; 0.990472,\; \backslash pm\; 0.103342,\; \backslash pm\; 0.091023)\; R\; .$$

Negating all vertices in both groups gives the mirror of the chiral snub dodecahedron, yet results in the same pentagonal hexecontahedron convex hull.

Isohedral variations can be constructed with pentagonal faces with 3 edge lengths.

This variation shown can be constructed by adding pyramids to 12 pentagonal faces and 20 triangular faces of a snub dodecahedron such that the new triangular faces are coparallel to other triangles and can be merged into the pentagon faces.

The pentagonal hexecontahedron has three symmetry positions, two on vertices, and one mid-edge.

File:Spherical pentagonal hexecontahedron.png

{{Icosahedral truncations}}

This polyhedron is topologically related as a part of sequence of polyhedra and tilings of pentagons with face configurations (V3.3.3.3.n). (The sequence progresses into tilings the hyperbolic plane to any n.) These face-transitive figures have (n32) rotational symmetry.

{{Snub table}}

• Truncated pentagonal hexecontahedron
• Amazon Spheres

{{Reflist}}

• {{The Geometrical Foundation of Natural Structure (book)}} (Section 3-9)
• {{Citation |last=Wenninger |first=Magnus |authorlink=Magnus Wenninger |title=Dual Models |publisher=Cambridge University Press |isbn=978-0-521-54325-5 |mr=730208 |year=1983 |doi=10.1017/CBO9780511569371}} (The thirteen semiregular convex polyhedra and their duals, Page 29, Pentagonal hexecontahedron)
• The Symmetries of Things 2008, John H. Conway, Heidi Burgiel, Chaim Goodman-Strauss, {{isbn|978-1-56881-220-5}} [https://web.archive.org/web/20100919143320/https://akpeters.com/product.asp?ProdCode=2205] (Chapter 21, Naming the Archimedean and Catalan polyhedra and tilings, page 287, pentagonal hexecontahedron )

• {{Mathworld2 |urlname=PentagonalHexecontahedron |title=Pentagonal hexecontahedron |urlname2=CatalanSolid |title2=Catalan solid}}
• [https://web.archive.org/web/20080924091620/http://polyhedra.org/poly/show/44/pentagonal_hexecontahedron Pentagonal Hexecontrahedron] – Interactive Polyhedron Model

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