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For example, in 2-space, a regular hexagon's edges all have the same magnitude, but also share a magnitude with the radius of the circumcentre (intuitive by sticking a six equilateral triangles together). In 3-space, you get a convex regular icosahedron (sticking 20 regular tetrahedra together at a mutual point).

So, is there a classification of these kinds of polyhedra where all the edges are the same length as the radius of the circumsphere?

joriki
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Mathgeek
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    That's not true. The solid angle subtended by a regular tetrahedron at one of its vertices is $\arccos\frac{23}{27}\approx0.551$ (see Wikipedia). That's not $\frac{4\pi}{20}=\frac\pi5\approx0.628$. I doubt that you'll find any regular (hyper)solids with this property other than the regular hexagon. – joriki Feb 28 '20 at 14:22
  • Indeed, the circumradius of the regular icosahedron is only about $0.951$ times the edge length. –  Feb 28 '20 at 14:23
  • Can an icosahedron not be constructed with several regular tetrahedra at a common corner point? Am I thinking of a different polyhedron? – Mathgeek Feb 28 '20 at 14:40
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    @Mathgeek. If you make accurate tetrahedra and join them together carefully to try and make an icosahedron, you will find that the result is a near-miss; the gaps do not quite close up. – Guy Inchbald Feb 28 '20 at 15:29
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    If you put 12 spheres around another sphere of the same size, they are not tightly packed. For a long time it was not known whether it was possible to fit a 13th sphere in there, as there seemed to be so much room, but it was proved only relatively recently that the Kissing Number in 3 dimensions is 12. If the 12 spheres did pack tightly in an icosahedral arrangement (around an equal 13th sphere), then 12 regular tetrahedra would build an icosahedron, but they don't. – Jaap Scherphuis Feb 28 '20 at 15:58

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The are no regular 3-space polyhedra for which this is true; the regular icosahedron is a near miss. The most symmetrical (and possibly the only?) example is the uniform or semiregular cuboctahedron (which, for this reason among others, Buckminster Fuller liked to call the Dymaxion solid).

Just as the hexagon can be constructed from equal-sided triangles, so the cuboctahedron can be constructed from equal-sided pyramids, with both square and triangular bases required.

See Richard Klitzing's answer for more about higher dimensions.

Guy Inchbald
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As was already stated in this answer for the acc. question "For which dimensions does it exist a regular n-polytope such that the distance of its vertex to its center equals the lenght of its side?" there is a whole dimensional series of Wythoffian polytopes, all of which have your asked for property. This very series sometimes is being described as expanded simplices.

But note that the tesseract and the 24-cell (both are 4D regular polytopes) would follow to your quest as well. That is, the above series by no means describes the only solutions. $-$ Further examples for instance could be found within the set of Gosset polytopes: $1_{22}$ within 6D, $2_{31}$ within 7D, and $4_{21}$ within 8D.

--- rk

Dr. Richard Klitzing
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