A piece of fabric shaped like a regular hexagon with unit-length sides is given, with vertices A...F and center O. Let a fold correspond to folding some fabric along a straight line, without stretching. For example, bringing A to E and B to D by folding along line FC is a fold. Let a uniform folding be a series of folds giving a result with uniform fabric thickness. For example, if a uniform-folding result is 3 layers thick somewhere, it is 3 layers thick everywhere.

Find a three-fold uniform folding giving a triangle.
Find a four-fold uniform folding giving a square rectangle.
Find a six-fold uniform folding giving a square, or prove that none exists.
Is there a five-fold uniform folding that gives a square?

Regular hexagon

  • $\begingroup$ Do all folds have to be along the labeled points? Can I do "fold the top quarter down so that AB is on the same line as FC"? $\endgroup$ – Kevin Aug 7 '14 at 15:39
  • $\begingroup$ @Kevin, yes, that fold's allowed. Folds don't need to include labeled points, which are there just for ease of reference for folds that involve corners or center. $\endgroup$ – James Waldby - jwpat7 Aug 7 '14 at 17:32

Here's a four fold uniform folding giving a square rectangle that is slightly wider than it is tall:

  • fold AB so that they lie on the line FC
  • fold ED so that they lie on the line FC

After doing that, A and E are in the same position, and B and D are also in the same position.


  • fold so that F lies on top of A and E
  • fold so that C lies on top of B and D

Thanks to PrisonMonkeys for correcting me on the square/rectangle nature of the result!

You can then obtain a 6-fold "square" by folding the 4-fold "square" above once vertically (down the middle), and then horizontally (down the middle), to obtain a "square" that is a quarter the size of the original.

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  • $\begingroup$ I'm pretty sure that the result is not a square, but a rectangle that is wider than it is high. $\endgroup$ – PrisonMonkeys Aug 7 '14 at 16:17
  • $\begingroup$ @PrisonMonkeys - oh man, you are absolutely right... thanks $\endgroup$ – jcdude Aug 7 '14 at 16:19
  • $\begingroup$ @jcdude, the altitude of a unit-side equilateral triangle is (√3)/2, or about .8660, so your result is about .8660 high, 1 wide. I originally wrote the title with the word rectangle instead of square, and with “giving a rectangle” instead of “giving a square”, but unfortunately changed it :( and will have to edit the question $\endgroup$ – James Waldby - jwpat7 Aug 7 '14 at 17:42
  • $\begingroup$ @jwpat7 thanks for the measurements! So I take it that there is a 4-fold solution that gives a perfect square? I can't see how to obtain such a solution so far... $\endgroup$ – jcdude Aug 7 '14 at 17:47
  • $\begingroup$ @jcdude, I don't know if there is, but doubt it (so, added “or prove that none exists” in question). The hexagon area is 3(√3) = √27, so a 2-thick UF square has area (√27)/2, or a side of √(√(27)/2), a 3-thick UF square has area √3, or a side of √√3, etc. I imagine it can be shown no finite UF can do that but don't know how to show it. BTW I thought of this problem while recently folding up my Eureka tent that has a hexagonal base. $\endgroup$ – James Waldby - jwpat7 Aug 7 '14 at 18:15

For a three fold giving a triangle of two thicknesses, fold along $AE, AC, CE$

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To get a triangle from a three-fold uniform folding, do the following:

  1. Bring A to E and B to D by folding along FC.
  2. Bring F to D by folding along EO.
  3. Bring C to E by folding along DO.

Now we have the triangle OED which is 6 layers thick everywhere.

I haven't found solutions to the other problems yet.

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