This shape is the first result what according to my judgement can be categorized as “archetypal”, which in short means something what reached a certain finalized state with some outstanding and well defined particularity.
At the same time it’s also kind of an “anomaly” as all the other previous designs are exclusively made of curved surfaces, while this one has only plain faces. No wonder that it was discovered by accident, while working on the level 4 continuity expansion.
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Generalities
A star polyhedron with tetrahedral symmetry, presenting 3D tessellation properties (honeycomb). It is composed of twenty four isosceles triangles, which are making up four sextet structures with three convex and three concave radial edges. It has ten vertices (four triple edge meetings, six quadruple edge meetings) and thirty-six edges (twenty-four convex and twenty-four concave). Every sextet is composed of three coplanar triangle pairs, whose constituents are touching in the vertices.
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The creation
1. The original method: I searched those points of a regular tetrahedron’s four symmetry axes (the lines joining the vertices and the face-centers), which are situated at equal distance from both the vertices of the faces they are belonging to and from the points situated somewhere on the lines joining the center with the edge midpoints (E in the below image). For the ABC face that point is O’.
It turned out that there is only a single possibility for my idea, namely when the formerly mentioned points are situated at sqrt(2)/2 distance from the center (if the edges are 1 unit). As the reach between the vertices and the center, the sqrt(3/8), coincides with the edge belonging 109.47 degree circle arcs top’s reach, I figured out that these points are corresponding to the edge belonging 141.057 degree ( 2*(180-109.47) ) circle arc top distances from the center.
Separately joining the six new points with all four vertices of the tetrahedron will give a structure composed of twenty-four isosceles triangles.
In the above image the 1 unit represents the radius of the circumsphere, thus the tetrahedron’s edges are 2*sqrt(2/3), approximately 1.633 in length. In the right figure the plane formed by the E, F, G points is closer, while the ABC triangle is farther from the observer. The O’ is situated halfway between them.
2. Starting from a tetrahedron: We construct four prisms from the faces of a regular tetrahedron, all with the height identical with the tetrahedron’s height. The composition made by the six double intersections of these prisms (this includes also the central quadruple intersection) is our solid.
3. Deducing from the rhombic dodecahedron: Four out of eight vertices represented by the triple edge meetings (one of the two tetrahedral compositions) of the mentioned catalan solid will be “sunk” twice as close toward the center. This is the easiest way for direct visualization and also tells about the close relationship between the two polyhedrons (more details in the final part).
The outcome of this maneuvre will be the disappearance of these vertices, as the new points will wind up on the lines joining the four unchanged triple edge meetings with the six quadruple edge meetings, namely on the edge midpoints of these (see the “original method”). The resulting twelve lines are creating four triple intersections, which are dividing the solid into twenty-four identical isosceles triangles.
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Formulas
Surface: S=12*sqrt(2)/ sqrt(3)*a^, around 9.7979*a^
Volume: V=8/3*sqrt(3)*a^3, around 1.5396*a^3
Surface in function of the circumscribed sphere: S=9*sqrt(2)/ sqrt(3)
Volume in function of the circumscribed sphere : V=r^3
V-sphere/ V-TetrTess=4/3*PI, around 4.18878
S-sphere/ S-TetrTess=4*sqrt(3)/ 9*sqrt(2) *PI, around 1.71006
(V-sphere/ V-TetrTess)/ (S-sphere/ S-TetrTess)=sqrt(6)
S-TetrTess/ V-TetrTess=9/sqrt(2), around 6.3639
One triangle: A=1; B,C=sqrt(11)/ sqrt(12), around 0.9574; alfa-angle: 62.965 degrees, beta-angle, gamma-angle: 58.518 degrees; S=sqrt(2)/ 2*sqrt(3)*a^
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The tessellation
During the joining of three individuals, one triangle of the first and one of the third solid will became coplanar forming a rhomb, whose characteristics are:
a=sqrt(11)/ sqrt(12); D=2*sqrt(2)/ sqrt(3); d=1; D/ d=2*sqrt(2)/ sqrt(3), around 1.633, the same as the ratio between the edge length and the radius of a circumscribed regular tetrahedron: 1/ 0.6123; A, C (acute) angles: 62.964 degrees; B, D (obtuse) angles: 117.036 degrees; S=sqrt(2)/ sqrt(3)*a^; the angle formed by two neighboring rhombs: 146.443 degrees.
Though similar, these rhombs are not the same as the ones which make up the rhombic dodecahedron. They are a little “thinner”, so if their small diagonals are the same, the ratio between the big diagonals is 2/sqrt(3), the same relation as the one between the sides and the height of an equilateral triangle.
Every vertex is shared by ten individuals (represented by six quadruple and four triple corners). Every edge is shared by six individuals. If this is vertical, than it’s the common part of three standing and three upside down solids, situated on the rotational axis of the sextet. Every individual is bordered by other four. If the central is a “standing” one, than all the surrounding four are “upside down” and vice versa. In other words: every individual can be “glued” only to the faces of the inversely positioned ones, thus the layers are forming chessboard-like 3D patterns. This can be illustrated well while using two different colors.
Inside the tessellation we can discern twelve non-parallel planes, represented by one triangle pair on every individual’s surface. The triangles are meeting in one vertex, forming a “hour-glass” shape, an individual is made up by twelve hour-glasses. To every plane belongs a rhombic pattern. The distance between two parallel rhombic rows belonging to the same plane is 4a, namely four times the small diameter of the rhombs, consequently the distance between two cell centers belonging to neighboring vertical rows is also 4a.
If we take into account only one “level” than we can observe that the individuals are joining each other alternating between the standing and upside down positions in a zigzag fashion, forming 120 degree angles with their centers.
This tessellation is heading towards a regular tetrahedron with infinitely long edges, during while the cells are forming bigger and bigger tetrahedral compositions, following the “square piramidal” sequence (1, 5, 14, 30, 55, 91…). Thus a “level four” structure is made up by thirty individuals, for example.
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Relatedness with regular polyhedra
If we join the triple edge meetings with the center, the solid can be divided into four identical parts, which if attached with their outer sides to the faces of a solid with similar size, their inverted inner sides together will form a rhombic dodecahedron.
This property can be found also in the case of the cube, where this inverting action results in six identical parts. The main difference between them is that the cube, the rhombic dodecahedron and the stellated rhombic dodecahedron (Escher’s solid) can turn into each other with the previously mentioned maneuvre during a cyclic process of three steps: cube, rhombic dodecahedron, Escher’s solid, cube, … and so on. In this “fractal sequence” the edge lengths of the similar solids (cubes for example) will successively double, while their volume will increase eight times.
The same process can be started from the tetrahedral tessellator, but there is “no return” to him. In turn, our shape can be constructed from the stellation of a certain kind of triakis tetrahedron. This is the shape represented by the quadruple (central) intersection of the four prisms described in the second method of the solid’s creation (see above). It can be also mentioned that if we circumscribe a tetrahedral tessellator inside a cube (the vertices tangent to the cube’s six faces) than its volume will be exactly one octa of the cube’s volume.
The further development of this shape lead me to the discovery of two other interesting solids: the contrastoid and the elementary tetrahedral tessellator.
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