I have seen this question on reddit (r/math's chatroom - to be specific). No one has answered it yet, so I thought I would pose the same question here.

The only discovery I made that is worth mentioning is this:

If it is possible to divide any triangle into convex pentagons then it is possible to divide the square as well (But not the other way around, so the triangle may as well be a waste of time).

Note: Just as I was writing this, an answer to the question was found (kind of?). But the answer is not very direct and there is still some confusion. Plus, I think this will pick the interest of people here as well. So I am going to ask it anyway.

Edit - Seems like that answer was actually wrong.


1 Answer 1


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We must find a face-subset of some pentagonal tiling such that at most four vertices are only incident to one pentagon only. If we only place three pentagons at a vertex, we get the dodecahedral tiling of a sphere. Any other configuration converges more slowly. We can remove at most four pentagons from the dodecahedron before we get too many vertices incident to only one face. This fixes the topology, and the rest is an exercise in embedding.


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  • 3
    $\begingroup$ Thank you! It seems very simple now that I have seen it, and I am kind of mad at myself for giving up and thinking it's impossible. $\endgroup$
    – Puzzlees
    Commented Jan 19, 2020 at 21:46
  • $\begingroup$ You can also divide a triangle into infinitely many pentagons by repeatedly slicing parallel to one edge. $\endgroup$ Commented Jan 20, 2020 at 8:19
  • $\begingroup$ Wouldn't there be a triangle at the top in that case, not a pentagon? $\endgroup$ Commented Jan 20, 2020 at 8:42
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    $\begingroup$ Seems like a potential for an OEIS entry: What is the minimum number of convex pentagons that can divide a convex N-gon? Starts with 0, 0, 9, 8, 1, 2, 2... $\endgroup$ Commented Jan 20, 2020 at 15:38
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    $\begingroup$ Hmm, actually, after a little experimentation, that sequence becomes not very interesting. f(6) = 2, just divide edge-to-edge. f(7) = 2, divide corner to edge, f(8) = 2, divide corner to corner, and after that it's just "Slice off pentagons until you can perform f(6-8) on the remaining bit, repeat, so it's just 2,2,2,3,3,3,4,4,4,5,5,5,6,6,6...etc. $\endgroup$ Commented Jan 20, 2020 at 15:52

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