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Cubiakis Icosahedron  
In 2008 I completed a polyhedral sculpture called MotherGlobe. When I exhibit this piece I am often asked "What IS that shape? Is there a name for it?" While there are couple of closely related polyhedra, I could not find any references to a polyhedron of this specific shape. While I expect that it has been the subject of analysis by many over the years, I have not been able to find any references that put a name to this specific polyhedra. Until I do I am calling it: Cubiakis Icosahedron. The Cubiakis Icosahedron is a irregular, nonconvex polygon composed of 60 identical faces each of which is an isosceles right triangle. This polygon is irregular because the faces do not have edges of the same length. It is nonconvex because there exits a line from two points inside the polygon that bisects the surface; for example a line between two tips would lie outside the polygon. Note: The HTML5/canvas animation below can be started and stopped with a double click. You can toggle the rendering from solid, points and wireframe by clicking the CTRL key.  
 
Note: you can spin the animation below by clicking the left mouse key, dragging the cursor across, then releasing the mouse key. Other LiveGraphics3D commands can be found here.  
 
The Cubiakis Icosahedron is similiar to the Small Triambic Icosahedron and Triakis Icosahedron but it is not among the 59 stellations of the icosahedron. To be considered a stellation of an icosahedron, each face must lie on the same plane as an original icosahedral face. That is not a property of the Cubiakis Icosahedron. This can be most easily seen by observing that the three faces surrounding each pyramid do not lie in the same plane, in contrast to the Small Triambic Icosahedron. The Cubiakis Icosahedron can be constructed by grafting twenty triangular pyramids onto an icosahedron where each face is an isosceles triangle with angles of 454590 degrees. The process of grafting polyhedrons onto faces of another polyhedron is called cumulation. Assuming the pyramids have a edge length $\alpha$, then the length of the base of the pyramid and the edge length of the icosahedron would have the same length $\alpha\sqrt{2}$. The height of this specific triangular pyramid will be $\frac{\alpha\sqrt{2}}{2}\sqrt{1Tan^2(\pi/6)}$ as shown below. Similar looking polyhedrons could be constructed by grafting triangular pyramids of various heights that range from very pointy to nearly flat.  
The distinguishing feature of the Cubiakis Icosahedron is that it can be constructed by dissecting 5 cubes and hinging pieces together. Each cube is dissected such that four of the eight corners of the cube become a point or vertex of a triangular pyramid. Each dissected triangular pyriamid is then hinged with another triangular pyriamid. The circumscribed tetrahedron is not used. The animation as well as the video clip below demonstrates how five dissected and hinged cubes can morph into a Cubiakis Icosahedron. Note: you can spin the animation below by clicking the left mouse key, dragging the cursor across, then releasing the mouse key. Other LiveGraphics3D commands can be found here.  
 
Dissecting and Hinging PolyhedraPeter Cromwell, in his 1997 book Polyhedra, explains that the Chinese developed mathematics around the third century BC. (page 24.) The Chinese used a process of dissecting polyhedra into known shapes so as to compute volume. Dissecting Polyhedra has a following within the puzzle making community. For example, consider Stewart Coffin's The Puzzling World of Polyhedral Dissections. Wolfram Research has several demonstrations of how polyhedrons can be dissected into other polyhedrons. For example, see Dissection of a Cube into Five Polyhedra. Not surprisingly, the prolific mathematician and sculptor George Hart also addresses polyhedral dissections. For example, he considers the dissection of the rhombic triacontahedron. Rubic's Snake is an example of dissected and polyhedra that pivot. However, the previous examples address dissections but do not involve hinging. Any consideration of dissected and hinged polygons or polyhedrons must consider the works of Greg N. Frederickson. Professor Frederickson has written numerous papers and three definitive books on this subject Hinged Dissections: Swinging and Twisting and Dissections Plane & Fancy and PianoHinged Dissections: Time to Fold!. Be sure to see his video gallery. The videos includes one by Daniel Wyllie and two videos of beautiful wooden models created by Walt van Ballegooijen. Daniel Wyllie also has a photo of another dissected and hinged polyhedra on Flickr and several interesting demonstrations on YouTube. Here is one called Icosahedron Inside Out and another called Rhombic Dodecahedron. Check them out. Robert Webb has a nice video showing a model of a regular dodecahedron morphing into a rhombic dodecahedron using hinged polyhedra. He calls his model Dodecamorph  The Dodecahedron ShapeShifter. More recent advances in the analysis of dissecting and hinging polyhedra is found in the computer science and computational geometry literature. For example, a paper by Erik D. Demaine, Martin Demaine, Jeffrey Lindy and Diane Souvaine "Hinged Dissection of Polypolyhdra" presents a "general family of 3D hinged dissections for polypolyhedra, i.e., connected 3D solids formed by joining several rigid copies of the same polyhedron along identical faces." See also Timothy G. Abbott, Abel, Charlton, E. Demaine, M. Demaine and Kominers Hinged dissections existThis citation has many references to this literature  


 
Properties of the Cubiakis Icosahedron
Circumradius: Minimum radius of a sphere that circumscribes the polygon. The circumradius of an cubiakis icosahedron is equal inradius of an icosahedron plus the height of a triangular pyramid consisting of right isosceles triangles with base length $\alpha\sqrt{2}$ and edge length. \[Inradius_{icosahedron}=\alpha (3\sqrt{3} +\sqrt{15})/12\] \[Inradius_{icosahedron}=\alpha 0.755761314\] The height of a triangular pyramid with edge length $\alpha$ and base length of $\alpha\sqrt{2}$ can be derived as follows. The base of the pyramid is an equilateral triangle with midradius equal to
\[Midradius_{equal lateral triangle} = \frac{\alpha\sqrt{2}}{2}Tan(\pi/6)\]
Therefore, the circumradius of a cubiakis icosahedron is equal to the inradius of an icosahedron plus the height of a pyramid of right isosceles triangles: \[Circumradius_{cubiakis icosahedron} = \alpha * 0.755761314 + \alpha 0.40824829\] \[Circumradius_{cubiakis icosahedron} = \alpha * 1.224744871\] The surface area of an cubiakis icosahedron is 5 times the surface area of a cube having edge length $\alpha$ \[Surface Area = 5*6\alpha^2 = 30\alpha^2\] The volume of an Cubiakis Icosahedron equals the volume of the encompassed icosahedron with edge length $\alpha\sqrt{2}$ plus the volume of the 20=5*4 triangular pyramids with edge length $\alpha$ and base length $\alpha\sqrt{2}$. Alternatively, one could add the volume of the five cubes minus the volume of the circumscribed tetrahedron that remains after cutting off the four triangular pyramids from the cube. (See tetrahedron or a derivation. Formulas for both are shown. (A) Volume of Cube less Volume of Circumscribed Tetrahedron \[Volume_{cube}=\alpha^3\] \[Volume_{tetrahedron}=(\alpha\sqrt{2})^3\frac{\sqrt{2}}{12}\] \[Volume_{(A)}=\alpha^3 (\alpha\sqrt{2})^3\frac{\sqrt{2}}{12}) \] \[Volume_{(A)}= \alpha^3*(1\frac{1}{3})\] \[Volume_{(A)}= \alpha^3 .666666666667\] (B) Volume of Four Triangular Pyramids with Right Triangle Faces \[Volume_{triangular pyramid}=\frac{1}{3}Bh\] \[B=Base_{equallateral triangle}=\frac{1}{2}(\alpha\sqrt{2})(\alpha\sqrt{2}Sin(\pi/3)) = \alpha^2Sin(\pi/3)\] \[h=Height_{right triangular pyramid} = \frac{\alpha\sqrt{2}}{2}\sqrt{1Tan^2(\pi/6)}\] \[Volume_{triangular pyramid}=\alpha^3*\frac{\sqrt{2}}{6}Sin(\pi/3)\sqrt{1Tan^2(\pi/6)}\] \[Volume_{triangular pyramid}= \alpha^3*.166666666667\] \[Volume_{(B)}= 4*\alpha^3*.166666666667 = \alpha^3*.666666666667\] (C) Volume of Cubiakis Icosahedron with edge length $\alpha$ \[Volume_{cubiakis icosahedron}=Volume_{icosahedron} + 5*Volume_{(A or B)}\] \[Volume_{icosahedron}=(\alpha\sqrt{2})^3\frac{5}{12}(3+\sqrt{5})\] \[Volume=\alpha^3 * (6.170765289 + 5*.66666667)\] \[Volume=\alpha^3 * 9.504098623\] There are two diheral angles in the cubiakis icosahedron. Along the edges of the right triangular pyramids which have length $\alpha$, the dihedral angle is $\frac{\pi}{2}$ or 90 degrees. Along the bases edges of length $\alpha\sqrt{2}$, the diheral angle is equal the dihedral angle of an icosahedron plus two times the face angle of a right triangular pyramid:
\[Height_{right triangular pyramid} = \frac{\alpha\sqrt{2}}{2}\sqrt{1Tan^2(\pi/6)}\]
\[Midradius_{equal lateral triangle} = \frac{\alpha\sqrt{2}}{2}Tan(\pi/6)}\]
\[FaceAngle_{equal lateral triangle} = Atan(\frac{\sqrt{1Tan^2(\pi/6)}}{Tan(\pi/6)})\]
\[FaceAngle_{equal lateral triangle} = 0.955316618 or 54.73561032^{\circ}\]
\[dihedral_{icosahedron} = cos^{1}(\sqrt{5}/3)\]
\[dihedral_{icosahedron} = 2.411864997 or 138.1896851^{\circ}\]
\[dihedral_{cubiakis icosahedron} = 2*0.955316618 + 2.411864997\]
\[dihedral_{cubiakis icosahedron} = 4.322498233 = 247.6609057^{\circ}\]
 
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