Longest shape name5/2/2024 The expansion is chosen so that the resulting rectangles are golden rectangles. The balls are "expanded" rhombicosidodecahedra, with the squares replaced by rectangles. The Zometool kits for making geodesic domes and other polyhedra use slotted balls as connectors. The rhombicosidodecahedron shares the vertex arrangement with the small stellated truncated dodecahedron, and with the uniform compounds of six or twelve pentagrammic prisms. If two bilunabirotundae are aligned this way on opposite sides of the rhombicosidodecahedron, then a cube can be put between the bilunabirotundae at the very center of the rhombicosidodecahedron. Two clusters of faces of the bilunabirotunda, the lunes (each lune featuring two triangles adjacent to opposite sides of one square), can be aligned with a congruent patch of faces on the rhombicosidodecahedron. Therefore, it has the same number of squares as five cubes. Therefore, it has the same number of triangles as an icosahedron and the same number of pentagons as a dodecahedron, with a square for each edge of either.Īlternatively, if you expand each of five cubes by moving the faces away from the origin the right amount and rotating each of the five 72° around so they are equidistant from each other, without changing the orientation or size of the faces, and patch the pentagonal and triangular holes in the result, you get a rhombicosidodecahedron. If you expand an icosahedron by moving the faces away from the origin the right amount, without changing the orientation or size of the faces, or do the same to its dual dodecahedron, and patch the square holes in the result, you get a rhombicosidodecahedron. There are different truncations of a rhombic triacontahedron into a topological rhombicosidodecahedron: Prominently its rectification (left), the one that creates the uniform solid (center), and the rectification of the dual icosidodecahedron (right), which is the core of the dual compound.įor a rhombicosidodecahedron with edge length a, its surface area and volume are:Ī = ( 30 + 5 3 + 3 25 + 10 5 ) a 2 ≈ 59.305 982 844 9 a 2 V = 60 + 29 5 3 a 3 ≈ 41.615 323 782 5 a 3 Geometric relations Johannes Kepler in Harmonices Mundi (1618) named this polyhedron a rhombicosidodecahedron, being short for truncated icosidodecahedral rhombus, with icosidodecahedral rhombus being his name for a rhombic triacontahedron.
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