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One can intersect a dodecahedron with a plane and obtain an equilateral triangle, a square, a regular pentagon, a regular hexagon, and a regular decagon:


         
         
          (Image of 6- and 10-gon from Mathworld.)


Q1. Does there exist a regular 7-gon, 8-gon, or 9-gon cross-section of the dodecahedron?

I can achieve, e.g., an irregular octagon, but not a regular octagon.

Q2. Can all five Platonic solids be achieved as cross-sections of one of the six regular 4-polytopes?

I haven't given this much thought, but the 120-cell seems the most likely candidate, as its facets are dodecahedra.

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  • $\begingroup$ I remember a related problem from a high school exam involving an octahedron. I suspect there is no regular polyhedron with a regular heptagonal or octahedronal cross section: one needs a pyramid or prism to achieve such. If I can, I will provide a sketch of justification later. $\endgroup$ Mar 30, 2014 at 18:40
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    $\begingroup$ In fact, I suspect the existence of slice S of polyhedron P implies G(S) is a subgroup (maybe even normal) of G(P). I am talking symmetry groups here. $\endgroup$ Mar 30, 2014 at 18:57
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    $\begingroup$ The square exists as a slice of a dodecahedron, but $D_8$ is not a subgroup of $C_2 \times A_5$. Specifically, the only order-$8$ subgroups of the latter are its Sylow $2$-subgroups, which are $C_2 \times C_2 \times C_2$. $\endgroup$ Mar 29, 2015 at 15:31

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