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I have been working on this question and I found that any regular polygon with n sides works.My claim is that we can cut any regular polygon of n sides into smaller regular polygons with n sides.And we will have smaller polygons with n sides and rhombuses.But the thing is that I haven't found a way to prove this.Is there anything wrong with my claim or is their a way to prove my claim?

Edit -This is for n=5 (not perfect) enter image description here

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    $\begingroup$ It would be good if you include diagrams of how you can do this, for several small values of $n.$ $\endgroup$
    – coffeemath
    Commented May 4, 2022 at 7:02
  • $\begingroup$ n=3 and n=4 is easy, but I'd definitely want to see what you're thinking of for n=5 or n=7. $\endgroup$
    – Jaap Scherphuis
    Commented May 4, 2022 at 8:37
  • $\begingroup$ Any equilateral polygon with an even number of sides and with opposite sides parallel (so that includes regular n-gons with n even) can be tiled by rhombuses. You can simply build a crescent strip of rhombuses along half the perimeter of the figure until it reaches the opposite edge to where you started. That leaves an (n-2)-gon that still has opposite sides parallel, so you can repeat this until the whole polygon is filled with rhombuses. This is a fairly well known result. You are doing something slightly different. .... $\endgroup$
    – Jaap Scherphuis
    Commented May 4, 2022 at 12:29
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    $\begingroup$ ... You are putting in two half-sized n-gons in there at opposite (or near opposite) corners, and then filling the two remaining areas with half-size rhombuses. Each of these area has an equal amount of boundary that is parallel, though not exactly on opposite sides, so it seems plausible that they can be filled with rhombuses too. Not sure of a proof though. $\endgroup$
    – Jaap Scherphuis
    Commented May 4, 2022 at 12:29

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