Wikipedia:Reference desk/Archives/Mathematics/2017 May 3

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May 3

How many Tetrahedra can touch at a point?

For Cubes, 8 can touch at a point, (filling all the space), for Octahedra 6 can touch (filling all the space) How many Tetrahedra can? (as followup, can you have more than 2 Dodecahedra or more than two Icosahedra?)Naraht (talk) 05:19, 3 May 2017 (UTC)

• The exact answer is "any positive number", using tetrahedra with a very "sharp" vertex, but you probably are talking about a regular tetrahedron (i.e. all faces are the same equilateral triangle). In which case, Tetrahedron#Formulas for a regular tetrahedron tells us the solid angle at a vertex is arccos(23/27). Since a full solid angle is 4π, that means you can have at most 4π/arccos(23/27) tetrahedra touching at a vertex, which is not an integer number; the same article tells us that Regular tetrahedra alone do not tessellate. Tigraan^{Click here to contact me} 12:09, 3 May 2017 (UTC)

• That gives me 22.8 which rounds down to 22. Doesn't that seem high ? Using the dihedral angle of 70.528779° I get 5 rotations about an edge, and since 3 edges meet at a point, that gives me 15, but 2 of those solids are shared between each pair of rotations, so I subtracted 5 to get 10. (Original, plus 4 rotations about first edge at vertex, plus 4 rotations about 2nd edge at vertex, minus shared solid, plus 4 rotations about 3rd edge at vertex, minus 2 shared solids.) Am I wrong, and, if so, were is my mistake ? StuRat (talk) 12:39, 3 May 2017 (UTC)

Clearly at least 20 is achievable. Decompose a regular icosahedron into twenty similar tetrahedrons, each of which is blunter-than-regular at the centre of the icosahedron. Then sharpen the tetrahedrons, so that they only touch at the centre. --catslash (talk) 14:44, 3 May 2017 (UTC)

My method may have missed some copies opposite the original solid, but that 22 figure still just seems too high. StuRat (talk) 15:02, 3 May 2017 (UTC)

This 2012 paper by Lagarias and Zong mentions the icosahedron argument (showing that the maximum is at least 20) and says that it is conjectured (but not proved) that 20 is indeed maximal. Gandalf61 (talk) 15:37, 3 May 2017 (UTC)

Your mistake lies in assuming that all the tetrahedra will share an edge with an "original" tetrahedron. In fact, your construction can at best cover half of the solid angle around the point (otherwise there would be a cross section through it where 3 triangles which were regular or "sharper" covered an angle greater than 180 degrees), so you could reflect it to give a lower bound of 20. MChesterMC (talk) 15:20, 4 May 2017 (UTC)

Six regular octahedra do not fill all the space around a vertex. One of the regular four-dimensional polytopes illustrates this. Another shows that 20 regular tetrahedra also have some wiggle room. —Tamfang (talk) 07:45, 9 May 2017 (UTC)

Yes, I meant regular tetrahedra. I wasn't sure that the tetrahedra made up of the center and a regular triangular face of the Icosahedra was "blunter" than a regular tetrahedra.Naraht (talk) 17:19, 3 May 2017 (UTC)

According to the regular icosahedron page, the radius of the inscribed sphere (and therefore the height of each of the constituent tetrahedra is) $\scriptstyle {{\frac {\sqrt {3}}{12}}(3+{\sqrt {5}})=0.755761}$ times the length of side. The tetrahedron page gives the height of a regular tetrahedron as $\scriptstyle {{\sqrt {\frac {2}{3}}}=0.816497}$ times the length of a side. Hence the tetrahedra in the icosahedron are relatively squat and therefore relatively blunt. Alternatively, since the icosahedron has fewer than 22.794665 sides, the solid angle at the apex of each of its constituent tetrahedra is greater than that of a regular tetrahedron. --catslash (talk) 22:08, 3 May 2017 (UTC)

Yes, but do we want 22.794665 sided objects? :) Thanx. The paper mentioned by Gandalf61 is a good start.Naraht (talk) 16:11, 4 May 2017 (UTC)

Dodecahedra and Icosahedra

For the other half of my original question, can 3 regular Icosahedra or 3 regular Dodecahedra meet at one point?Naraht (talk) 16:11, 4 May 2017 (UTC)

This amounts to asking if the corresponding dihedral angles are less than 120 degrees. According to Table of polyhedron dihedral angles, it's 116 degrees for the dodecahedron, so the answer is yes in that case. It's 138 degrees for the regular icosahedron so the answer is no there. But for the rhombic dodecahedron the angle is exactly 120 degrees, which accounts for the Rhombic dodecahedral honeycomb. --RDBury (talk) 20:30, 4 May 2017 (UTC)

standupmaths has a video in which dodecahedrons are stitched together: https://www.youtube.com/watch?v=H18pyUN4U1M (fast forward to about 7:00) Best wishes Robinh (talk) 21:31, 4 May 2017 (UTC)

Awesome! And the idea that the rhombus turns out as a golden ratio rhombus is quite interesting, but if you toss around enough fives, you seem to get there a lot...Naraht (talk) 14:04, 7 May 2017 (UTC)

One of the regular four-dimensional polytopes is built of dodecahedra meeting three to an edge. To make three regular icosahedra meet at an edge, you have to go into hyperbolic 3-space. —Tamfang (talk) 07:45, 9 May 2017 (UTC)

...which creates the icosahedral honeycomb. It is also possible to make more than 3 icosahedra meet at a point in hyperbolic 3-space, but the result will have to spill over to the ultra-ideal region beyond infinity in hyperbolic projective space, corresponding to the outside of the Poincaré ball model. Double sharp (talk) 07:52, 9 May 2017 (UTC)