Apr 24
Pentagonal Hexecontahedron
z_(n+1) = z_n^(6.5 + 9* |t - 0.5|) + i^(-4t) * c
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Pentagonal Hexecontahedron
z_(n+1) = z_n^(6.5 + 9* |t - 0.5|) + i^(-4t) * c
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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).
Regular-ish Convex Polyhedra Bracket — Round 3
Which shape is better?
Regular Tetrahedron
Rhombic Triacontahedron
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
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)
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
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Disdyakis Dodecahedron
z_(n+1) = 0.5 * ((1 + cos(2πt)) * z_n^2.5 - (1 - cos(2πt)) * z_n^3.5) + i^(4t)c
Deltoidal Icositetrahedron
z_(n+1) = z_n^(7 - 4 * |t - 0.5|) + kc
k is 1 if t < 0.5; -1 otherwise.
Triakis Octahedron
z_(n+1) = 0.5 * (i^(4t+2) * (1 + cos(2πt)) * z_n⁵ + i^(-4t)(1 - cos(2πt)) * z_n⁷) + c