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How many space that divided by faces of regular convex polyhedron (without interior area)?

If the case is tetrahedron, We can easily imagine total $4+6+4=14$ space that faces of tetrahedron devided.

And case of cube, $8+9+9=26$ space will devided.

But what happened in octahedron.

I imagine $25+4$in down side, symmetry in upwards. So $58$ is my answer but I can't easily trust myself... How to I get right answer in logical way??

user366725
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  • It seems like you are trying to ask, "Into how many regions is $3$-space divided by the planes containing the faces of a regular polyhedron?" If that is so, I don't quite see how you get your answer for a cube. Are you not counting the interior of the cube? – saulspatz Aug 21 '19 at 14:15
  • I fully agree with the formulation by saulspatz. Besides, you should begin by the same question about a formula for the regular polygon with $n$ sides in 2D. – Jean Marie Aug 21 '19 at 14:37
  • @saulspatz It is intuition. But i could say $8+12+6=26$ – user366725 Aug 21 '19 at 14:39
  • You should add $1$ for the interior of the cube, making $27=3 \times 3 \times 3$ as is logical when reasoning by horizontal cuts. – Jean Marie Aug 21 '19 at 14:42
  • That doesn't really answer my question. Are you disregarding the interior of the cube intentionally? – saulspatz Aug 21 '19 at 14:42
  • @saulspatz Yes, I don't count interior area. I'll fix it~ – user366725 Aug 21 '19 at 14:43
  • Are you aware of the general formula https://math.stackexchange.com/q/2312255 (take D=3) ? for $n$ planes ? – Jean Marie Aug 21 '19 at 14:50
  • @JeanMarie polyhedron's faces cut spaces maximum??? – user366725 Aug 21 '19 at 14:54
  • No, far from this maximum number because many plane couples are parallel in an octahedron... – Jean Marie Aug 21 '19 at 14:56
  • Look at the highest-voted answer to https://math.stackexchange.com/questions/1911252/greatest-number-of-parts-in-which-n-planes-can-divide-the-space This doesn't solve your problem, because you have parallel planes and planes meeting in more than $3$ points. You might try to see if you can adjust the argument for these cases. Start in two dimensions. – saulspatz Aug 21 '19 at 16:31
  • @saulspatz Thank you for answering my deficient question! I'll try it – user366725 Aug 21 '19 at 16:33

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Note: The analysis below is not accurate for the dodecahedron and icosahedron, both of which have regions that do not contact the polyhedron at all. But I decided to leave it up because it does work for the other 3 regular polyhedrons.

Every exterior region contacts the polyhedron in a face, edge or vertex. Further, for each face, edge, or vertex, there is only one exterior region that contacts it and not some larger element with it on the boundary.

Therefore the total number of regions is $1 + V + E + F$,

  • $1$ for the interior of the polyhedron,
  • $V$ for all the exterior regions that only contact a vertex,
  • $E$ for all the exterior regions that contact an edge, but not a face.
  • $F$ for all the exterior regions that contact a face.
Paul Sinclair
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  • +1 for the nice analysis - even though it's not certain that this is the OP's question. – Ethan Bolker Aug 21 '19 at 22:03
  • @EthanBolker - (s)he seems to agree with saulspatz's interpretation, which also appeared likely to me. I tend to be more forgiving when someone is obviously not proficient in English, being well aware of how proficient I am in any other language. – Paul Sinclair Aug 21 '19 at 22:54
  • The planes of an icosahedron define ten kinds of bounded external region, not all of which touch the core icosahedron, even at a vertex. – Anton Sherwood Aug 22 '19 at 06:28
  • @AntonSherwood - I only see one type of bounded external region. However, you are right that some of the unbounded external regions do not touch the icosahedron at all. And the same is true of the dodecahedron. – Paul Sinclair Aug 22 '19 at 14:33
  • See https://en.wikipedia.org/wiki/The_Fifty-Nine_Icosahedra … and the Talk page, answering my question, points to a diagram showing seven kinds of unbounded cell, none of which touch the core. – Anton Sherwood Aug 23 '19 at 06:18
  • @AntonSherwood - I'm sorry my last response was confusing. The only thing in your previous comment I disagreed with was the word "bounded", which appears to have been an accidental inclusion. The only reason I didn't completely delete my erroneous analysis was that it does work when the exterior angles between adjacent faces is >270 degrees. So I think it is still helpful as a partial answer – Paul Sinclair Aug 23 '19 at 16:08
  • The word ‘bounded’ was intended. Where are you looking, that you see only one type? – Anton Sherwood Aug 28 '19 at 22:21
  • @AntonSherwood - Just inadequate visualization on my part. I was failing to consider planes that didn't immediately border a vertex, but still "lean in" towards it. And some similar cases. – Paul Sinclair Aug 29 '19 at 16:17