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We're going to take the 5 platonic solids (tetrahedron, cube, octahedron, dodecahedron, and icosahedron) and suspend them in various ways (we'll assume that they are solid and of uniform density). Then we'll do a horizontal cut through the centre of gravity and describe the shape of the resulting cut face.

The suspension methods will be:

  1. By a vertex (so any one of the vertices)
  2. By a face (meaning by the middle of one of the faces)
  3. By an edge (meaning the middle of one of the edges)

(This is a generalisation of a problem that I remember describing to a group, and only convincing them that the answer I gave was right by cutting up a potato. It was the cube, vertex case)

I've given some of the simpler answers in brackets, as examples. Bonus points for doing it all in your head...

  1. Tetrahedron suspended by a vertex. (equilateral triangle)
  2. Tetrahedron suspended by a face. (equilateral triangle)
  3. Tetrahedron suspended by an edge.
  4. Cube suspended by a vertex.
  5. Cube suspended by a face. (square)
  6. Cube suspended by an edge. (rectangle)
  7. Octahedron suspended by a vertex. (square)
  8. Octahedron suspended by a face.
  9. Octahedron suspended by an edge.
  10. Dodecahedron suspended by a vertex.
  11. Dodecahedron suspended by a face.
  12. Dodecahedron suspended by an edge.
  13. Icosahedron suspended by a vertex.
  14. Icosahedron suspended by a face.
  15. Icosahedron suspended by an edge.

There's a couple in there that get a little tricky!! Enjoy!

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  • $\begingroup$ Hmm, with nothing restricting the answer, a general answer for all would be a closed polygon, wouldn't it? Still describing the shape ;c) $\endgroup$
    – BmyGuest
    Commented Aug 26, 2015 at 6:32

2 Answers 2

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  • 3: square (cuts across all four faces symmetrically)
  • 4: regular hexagon (cuts across all six faces symmetrically)
  • 8: regular hexagon (cuts across six faces symmetrically)
  • 9: rhombus (passes through two vertices and across two edges, cutting four faces in half)
  • 10: regular hexagon (cuts across six faces symmetrically)
  • 11: regular decagon (cuts across ten faces symmetrically)
  • 12: irregular hexagon (passes along two edges and across two edges, cutting four faces in half)
  • 13: regular decagon (cuts across ten faces symmetrically)
  • 14: regular hexagon (cuts across six faces symmetrically)
  • 15: irregular hexagon (passes along two edges and across two edges, cutting four faces in half)

The cross-sections for each polyhedron suspended from a vertex are the same as for its dual suspended from a face. The edge cross-sections are less regular because the orientation of the edge breaks rotational symmetries.

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  • $\begingroup$ Number 14 is not a hexagon. $\endgroup$
    – Florian F
    Commented Jul 7, 2020 at 7:30
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  1. Equilateral triangle
  2. Equilateral triangle
  3. Square
  4. Regular hexagon
  5. Square
  6. Rectangle
  7. Square
  8. Regular hexagon
  9. Parallelogram
  10. Regular hexagon
  11. Regular decagon
  12. Regular hexagon
  13. Regular decagon
  14. Regular dodecagon
  15. Oblique hexagon
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  • $\begingroup$ Any explanation??? $\endgroup$
    – Rohcana
    Commented Aug 26, 2015 at 1:22
  • $\begingroup$ Not all correct, I think. Not that I'm necessarily an authority $\endgroup$
    – Dr Xorile
    Commented Aug 26, 2015 at 1:22
  • $\begingroup$ I mean 10, not 9. Sorry. $\endgroup$
    – Dr Xorile
    Commented Aug 26, 2015 at 2:03
  • $\begingroup$ @Anachor I don't know what kind of explanation you are expecting. It's not a riddle or a strategy question, where you can explain your reasoning. For this question, that's just the way it is. No kind of explanation can change the answer or further clarify it. $\endgroup$
    – Illyasviel
    Commented Aug 26, 2015 at 19:21
  • $\begingroup$ Look at f" answer, I'm asking how you found those results, by working it your head, or by a software or by google, If you did it in your head, how did you visualise it, etc. $\endgroup$
    – Rohcana
    Commented Aug 27, 2015 at 1:07

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