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Dec 15, 2022 at 2:22 history edited Oscar Lanzi CC BY-SA 4.0
added 1 character in body
Dec 15, 2022 at 0:36 history edited Oscar Lanzi CC BY-SA 4.0
Added recursion rules.
Dec 14, 2022 at 23:54 comment added Oscar Lanzi Or, set off a triangle along each edge and leave a smaller pentagon 8n the center. My 7+11 iteration is too busy to present, but it ends up doing just tgwt with each piece further subdivided into three ($6×3=18=7+11$).
Dec 14, 2022 at 23:49 history edited Oscar Lanzi CC BY-SA 4.0
Modified Pic. One ogmf my divisions was inconsistent.
Dec 14, 2022 at 22:32 comment added Oscar Lanzi 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.
Dec 14, 2022 at 22:11 comment added Per Alexandersson 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?"
Dec 14, 2022 at 17:25 history edited Oscar Lanzi CC BY-SA 4.0
edited body
Oct 11, 2022 at 20:32 comment added Wlod AA 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).
Oct 11, 2022 at 20:23 comment added Oscar Lanzi In the divisions shown here, the triangles are isosceles. So mirror reflection is trivial.
Oct 11, 2022 at 20:12 comment added Wlod AA 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).
Oct 11, 2022 at 17:49 history edited Oscar Lanzi CC BY-SA 4.0
edited body
Oct 11, 2022 at 16:56 history answered Oscar Lanzi CC BY-SA 4.0