Say that a 3D shadow of a 4-polytope is a parallel projection to 3-space, not necessarily orthogonal to that 3-space (that would make it an orthogonal projection). I am wondering if each of the five regular polyhedra in 3D are shadows of regular 4-polytopes. The 4-simplex can project to a regular tetrahedron, the tesseract to a cube, the 16-cell can parallel project to a cube or a regular octahedron. But I do not know if the 120-cell can project to a regular dodecahedron, or if the 600-cell can project to a regular icosahedron. (I think not?) Perhaps under perspective projection they can? Failing that, perhaps irregular convex 4-polytopes have the regular dodecahedron and icosahedron as shadows?

I am not as familiar with the regular 4-polytopes as I should be; otherwise the answers would be evident. I am sure those more knowledgeable can answer my question easily. Thanks!

Addendum. The confusion in my question (sorry!) caused a confusing welter of comments (some now deleted), but I think matters are clearer now. Three of the Platonic solids can be achieved as parallel-projection shadows (light at $\infty$): tetrahedron, cube, octahedron. As Theo explains in his answer, the dodecahedron and icosahedron cannot be so achieved. However, the dodecahedron can be achieved as a perspective-projection shadow (light at a finite point). The analogy with 3D is as follows. If a light is placed close to a face of the dodecahedron, a pentagon shadow results. Similarly, if a light is placed close to dodecahedral facet of the 120-cell, a dodecahedral shadow results (as Theo says).

What remains unclear to me is if the icosahedron can be achieved as a shadow. If one could place the light directly on a vertex of the 600-cell, then the shadow should be its vertex figure, which is an icosahedron. But if the light is exterior to the polytope, the shadow is more complicated.

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    $\begingroup$ I don't see the logic of this question. In 2-dim, how is a regular pentagon a projection of a regular 3-dim polytope (say icosahedron would be a natural candidate)? Similarly, why would you expect an icosahedron to be a projection of a 600-cell? Not at all! $\endgroup$
    – Igor Pak
    Nov 10, 2010 at 1:27
  • $\begingroup$ A pentagonal prism has a regular pentagon as a projection. I suppose there is some (irregular, convex) 4-polytope that is to the regular dodecahedron or icosahedron as the pentagonal prism is to the pentagon. $\endgroup$ Nov 10, 2010 at 1:34
  • $\begingroup$ georgehart.com/rp/120-cell.gif $\endgroup$
    – Ian Agol
    Nov 10, 2010 at 1:36
  • $\begingroup$ @Igor: Perhaps you are right, there is little logic here. However, a regular pentagon is a shadow of a dodecahedron from a light just above one face. @Agol: Yes, that is a beautiful construction! But I am unclear if it is a "shadow." $\endgroup$ Nov 10, 2010 at 1:41
  • $\begingroup$ @Igor: I was imagining three types of shadow projection: orthogonal, parallel, and perspective, with the latter corresponding to light at a point. And wondering which is needed to obtain the Platonic solids. So your answer that the regular icosahedron is a perspective shadow but not a parallel shadow is what I was seeking. So: Thanks! And apologies for the lack of clarity. $\endgroup$ Nov 10, 2010 at 1:59

2 Answers 2


Joseph, in your response to Igor in the comments you allow holding the candle that's casting the shadow very close to a face. This is precisely the picture that Ian gave in the comments. But in your question you require that the candle be infinitely far away, so that it is a parallel projection. If you do allow holding the candle very close to a face, then it suffices to find a 4-tope that has your given 3-tope as a face, and the 600-cell has 20-cell faces and the 120-cell has 12-cell faces.

But if you require your candle to be infinitely far away, I think you're hosed. From three to two, the face-centered shadows of the dodecahedron and icosahedron are, respectively, the regular 10- and 6-gons, for example. Similarly, any parallel shadow of the 120 or 600 cell will look like some sort of "equator", which you can see will have many too many facets to be an icosahedron or a dodecahedron. For example, here's a parallel projection orthogonal to a face of the 120 cell.

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    $\begingroup$ The 600-cell has 4-cell (tetrahedron) faces. $\endgroup$ Nov 10, 2010 at 2:52
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    $\begingroup$ Who is Ian, and where is his comment? Has that, too, fallen into a shadow? $\endgroup$ Nov 10, 2010 at 5:06
  • $\begingroup$ @Theo: As Tom mentions, I think you meant that the vertex figure of a 600-cell is a 20-face cell (in fact, an icosahedron). $\endgroup$ Nov 10, 2010 at 12:16
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    $\begingroup$ @Gerry: I believe "Ian" = "Ian Agol" = "mathoverflow.net/users/1345/agol". $\endgroup$ Nov 10, 2010 at 15:21
  • $\begingroup$ @Tom: Oops, yes. Anyway, some shadow of the 600 cell is the 20-cell --- the shadow that Wikipedia uses for its picture on "600 cell". But this is a shadow with the candle close to the object. @Gerry: Sorry, I meant Ian Agol's comment to the original post; my answer started out as a comment. $\endgroup$ Nov 11, 2010 at 2:37

Just for fun, here's a non-answer all about the case of 3 projected to 2. i never thought all of this through before. For each of the five Platonic solids you can ask what do you see if you look at it from a face direction or from a corner direction. The answer usually depends on how far away it is. (I'm thinking of an eyeball instead of light source, so "what you see from there" rather than "what the shadow looks like".) Taking the obvious ones first, we have:

4-hedron face or corner view: Always a regular 3-gon

6-hedron face view or 8-hedron corner view: Always a regular 4-gon

8-hedron face view: From infinitely far, a regular 6-gon. From large finite distance, a 6-gon with all sides equal but only regular 3-gon symmetry. At a certain distance (namely where you suddenly can see only 1 face rather than 1+3) a regular 3-gon, and so it remains until it hits you like a cream pie.

6-hedron corner view: Ditto, except that the 6-gon doesn't turn into a 3-gon until the final moment when it pokes you in the eye.

12-hedron face view: Exactly like the 8-hedron face view, except change 6 to 10 and 3 to 5.

20-hedron corner view: Just like 6-hedron corner view ($\times \frac{5}{3}$) except that the 10-gon does not wait until the final moment to become a 5-gon.

20-hedron face view: Always has at least the symmetry of a regular 3-gon. Initially a regular 6-gon, then an irregular 6-gon with equal sides, then suddenly the angles reverse their direction of change. Passes through a regular 6-gon again, then back to an irregular 6-gon and eventually turns into a regular 3-sided cream pie.

12-hedron corner view: Always has at least the symmetry of a regular 3-gon. Initially a 12-gon with the symmetry of a regular 6-gon, equal angles but not equal sides. Then an irregular 12-gon with merely 3-gon symmetry. For one moment a 6-gon (not regular, I'm sure) then an irregular 9-gon until it pokes you in the eye.

I don't know what the corresponding story is for the thing whose faces are 600 4-hedra, or the thing whose faces are 120 12-hedra. I started thinking about the thing whose faces are 24 8-hedra.

  • $\begingroup$ Ha! I like the "pokes you in the eye" characterization! The hexagon from one point above a vertex of a dodecahedron is interesting. I believe its edge lengths are in the golden ratio: three =1, three =1.618? $\endgroup$ Nov 10, 2010 at 16:22
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    $\begingroup$ Cream pies, eye pokes... is this mathematics or a Three Stooges short? $\endgroup$ Nov 10, 2010 at 18:52
  • $\begingroup$ I think it's what Reader's Digest would call "More Picturesque Speech". Obviously don't try it at home without protective goggles. $\endgroup$ Nov 10, 2010 at 19:09

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