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, ocatahedron.
As Theo explains in his answer, the dodecahedron and icosahedron cannot be so achieved.
However, the dodecahedron can be achieved as a perpsective-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.