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I posted this on math.stackexchange, but got no answers.

It is easy to divide a 2-gon into 3 congruent line segments. It is also easy to divide a triangle into 4 smaller triangles that are congruent. One of Martin Gardner's favorite problems (as he writes in one of his books) is to show that one can divide a square (regular 4-gon) into five congruent and connected pieces.

The natural question is then: can one subdivide a regular pentagon into six congruent connected pieces?

This sounds related to Monsky's theorem.

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    $\begingroup$ I'm not sure it is the natural question (although it is a good one, I'd guess no.) The first fact concerns the $1$ dimensional simplex which is also the $1$ dimensional hypercube and cross-polytope. A regular $n$-gon can be dissected in may ways into $n$ congruent pieces. An arbitrary triangle can be naturally dissected into $4$ congruent triangular pieces. Seeing the natural ways to dissect a rectangle into $4$-pieces it is an aha moment to see how to get $5.$ The natural question (which might not be that hard) is about simplices and boxes in $3$ and more dimensions. $\endgroup$
    – Aaron Meyerowitz
    Commented Oct 31, 2016 at 5:56
  • $\begingroup$ There must be some further condition than congruent and connected--why not just divide one axis of the square evenly into $n$ segments along one axis and cross them with the other axis? Voila, $n$ congruent and connected pieces. $\endgroup$
    – Steve
    Commented Nov 2, 2016 at 1:07
  • $\begingroup$ @Steve I do not follow your solution - it is about subdividing the pentagon, not the square. The square, as you say, is quite easy to divide into any number of congruent pieces. $\endgroup$
    – Per Alexandersson
    Commented Nov 2, 2016 at 1:47
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    $\begingroup$ I think Gardners point was that here are a plethora of ways to get $4$ congruent pieces, for example a path from the center to a side and its rotations by $90$ degrees. Somehow, knowing that makes it hard to see the obvious (and perhaps only) solution for $5.$ The fun is the aha moment. $\endgroup$
    – Aaron Meyerowitz
    Commented Jun 20, 2018 at 21:43
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    $\begingroup$ @AaronMeyerowitz, Gardner used a misdirection. Otherwise, there is no "aha" moment about getting n congruent pieces, it'd be totally and instantly obvious to about everybody. $\endgroup$
    – Wlod AA
    Commented Oct 11, 2022 at 20:22

1 Answer 1

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If we allow two sets of congruent pieces, then the regular pentagon can divided into various numbers of parts according to the Lucas numbers. Usually these are defined by the recursion $L_1=1,L_2=3, L_{n+1}=L_n+L_{n-1}$ for $n\ge2$. If we run the recursion backwards, we find $L_0=2$ and $l_{-n}=(-1)^nL_n$. We observe the (positive) numbers from this sequence in the divisions below, where the two sets of congruent areas are represented by different colored dots. With an unbounded number of iterations the ratio of the numbers of pieces tends to $\phi=(1+\sqrt5)/2$, and the nodes fill in a tenfold symmetric quasilattice.

enter image description here

The rules for the divisions are as follows:

  1. Start by drawing two diagonals of the pentagon. This produces two shapes of isosceles triangles, one acute and one obtuse. Each shape will be reproduced with smaller size and greater count of triangles by the subsequent divisions governed by Rules 2 and 3.

  2. When the acute triangles are larger, subdivide them by bisecting one of the base angles. The base angle to be bisected is chosen as follows: rotate the figure so each acute triangle in turn has the apex on top, then select the base angle at the bottom left of the triangle.

  3. When the obtuse triangles are larger, subdivide them by drawing one trisector of the apex angle. The trisector to be drawn is chosen by rotating the fugure so that each obtuse triangle in turn has its apex on top, then drawing the trisector that os directed downwards and to the right.

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  • $\begingroup$ It's very simple to divide the regular pentagon into 6 pieces of 2 isometric types. However, the way I see it, the isometry of two pieces requires an orientation change (mirror symmetry). It may be difficult to avoid this (I don't know). $\endgroup$
    – Wlod AA
    Commented Oct 11, 2022 at 20:12
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    $\begingroup$ In the divisions shown here, the triangles are isosceles. So mirror reflection is trivial. $\endgroup$
    – Oscar Lanzi
    Commented Oct 11, 2022 at 20:23
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    $\begingroup$ However, the OP's q. was about 6 pieces. Two different pieces is already taking liberties. (Nevertheless, your answer is interesting, I've learned something that was new to me). $\endgroup$
    – Wlod AA
    Commented Oct 11, 2022 at 20:32
  • $\begingroup$ So, a pentagon into 6 pieces, of two different congruence classes, is also difficult it seems. Sounds like a generalization of the problem, "a regular N-gon into k-pieces, which belong to C congruence classes, is it doable or not?" $\endgroup$
    – Per Alexandersson
    Commented Dec 14, 2022 at 22:11
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    $\begingroup$ I chose the Lucas number based division for number theoretical and geometrical elegance. If you want six pieces, in a (to me) less elegant fashion, bisect the isosceles triangles in my 2+1 division. $\endgroup$
    – Oscar Lanzi
    Commented Dec 14, 2022 at 22:32

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