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Inspired by this question. Same challenge as the original, but with a regular pentagon instead!

Copied below for reference (with changes to shape and numbers):

How can you dissect a regular pentagon into five non-self-intersecting congruent polygons such that each polygon has the same perimeter length as the given regular pentagon?

The goal is to have minimal number of edges per polygon.

Bonus question: Find a numerical representation of the pentagon and polygons in addition to a theoretical dissection.

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  • $\begingroup$ What does "an integer representation of the shape" mean? $\endgroup$ Commented May 13 at 18:43
  • $\begingroup$ I guess it means all line lengths are integers $\endgroup$ Commented May 14 at 2:58
  • $\begingroup$ I guess I meant more like "Find a numerical representation of the pentagon and polygons in addition to a theoretical dissection" $\endgroup$ Commented May 14 at 17:04

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I don't think this can be done with

fewer than seven sides per polygon.

I don't have any mathematical proof, but it seems that you just can't get sufficient perimiter without at least

two concave corners,

and that doesn't seem possible without at least

seven sides.

Here is my attempt at a solution that meets my criteria above, following a similar pattern to the solution to the linked question.
I haven't measured the perimeter, but it is evident that by adjusting the side lengths, it should be trivial to achieve a perimeter equal to that of the pentagon.

A visual solution to the problem, consisting of five rotationally-symmetric septagons, each with two concave angles, nested within each other

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    $\begingroup$ GeoGebra One point is free to move on the pentagon, the other is constrained to an ellipse. $\endgroup$ Commented May 13 at 19:17
  • $\begingroup$ @DanielMathias That's a really cool demonstration. Thanks! Are you saying one point is constrained to an ellipse if you want to keep the perimeter the same? $\endgroup$ Commented May 13 at 21:36
  • $\begingroup$ Yes, that is correct. $\endgroup$ Commented May 13 at 22:35
  • $\begingroup$ @GentlePurpleRain en.wikipedia.org/wiki/Ellipse#Definition_as_locus_of_points $\endgroup$ Commented May 14 at 8:17
  • $\begingroup$ In my post I used two thumbtacks on a cardboard and a thin wire with same length as square perimeter with a pencil to find the solution before posting it. $\endgroup$ Commented May 14 at 8:19

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