#catalan solids

[vedalkensamurai]

Pentagonal Hexecontahedron

z_(n+1) = z_n^(6.5 + 9* |t - 0.5|) + i^(-4t) * c

[art-of-mathematics]

Only 4 more catalan solids - and then we will have drawn them allll....

^ Catalan solids and their duals, the Archimedean solids, are depicted here. The same position of each of the 13 solids per paper sheet depicts the dually relatd solids, i. e.: The truncated tetrahedron's dual is the triakis tetrahedron (both are located in the upper right corner of the paper sheets).

[best-shapes]

Regular-ish Convex Polyhedra Bracket — Round 3

Which shape is better?

Regular Tetrahedron

Rhombic Triacontahedron

Voting ended onJan 4, 2024

Propaganda

Regular Tetrahedron:

Also called the Triangular Pyramid

Platonic Solid

Regular

Dual of the Regular Tetrahedron

It has 4 regular triangular faces, 6 edges, and 4 vertices.

Self-Dual

Image Credit: Cyp

Rhombic Triacontahedron:

Also called the Triacontahedron

Catalan Solid

Dual of a quasiregular polyhedron

Dual of the Icosidodecahedron

It has 30 rhombic faces, 62 edges, and 32 vertices of two types.

One of the 9 edge-transitive convex polyhedra along with the 5 Platonic Solids, the 2 Quasiregular Convex Polyhedra, and the Rhombic Dodecahedron.

Image Credit: Maxim Razin based on Cyp

[knotty-et-al]

New polyhedron drawing just dropped:

Triakis Icosahedron [dual of the Truncated Dodecahedron]

In order to draw that triakis icosahedron I drew a truncated dodecahedron (picture of truncated dodecahedron attached above)

[vedalkensamurai]

Deltoidal Hexecontahedron

z_(n+1) = 0.5 * (i^(4t) * (1 + cos(2πt)) * z_n^5.5 + i^(-6t)(1 - cos(2πt)) * z_n^3.5) + c

[vedalkensamurai]

Disdyakis Dodecahedron

z_(n+1) = 0.5 * ((1 + cos(2πt)) * z_n^2.5 - (1 - cos(2πt)) * z_n^3.5) + i^(4t)c

[vedalkensamurai]

Deltoidal Icositetrahedron

z_(n+1) = z_n^(7 - 4 * |t - 0.5|) + kc

k is 1 if t < 0.5; -1 otherwise.

[vedalkensamurai]

Triakis Octahedron

z_(n+1) = 0.5 * (i^(4t+2) * (1 + cos(2πt)) * z_n⁵ + i^(-4t)(1 - cos(2πt)) * z_n⁷) + c