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Q. Is there a tiling of the plane by one each of simple polygons of $n$ vertices: one triangle, one quadrilateral, one pentagon, $\ldots$ , one simple polygon of $n$ vertices, $\ldots$ ?

Here a vertex of a polygon is a point on its boundary with internal angle that differs from $\pi$.

I suspect the answer to Q is Yes but I am not seeing an explicit construction.

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    $\begingroup$ If the polygons don't have to be convex then you can just tile the plane with pentagons and divide each pentagon into two polygons of order n and n+1 (n=3,5,7,9,...) with a polygonal path. $\endgroup$
    – Noam D. Elkies
    Commented Nov 19 at 16:26
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    $\begingroup$ @NoamD.Elkies: Brilliant---Thanks! $\endgroup$
    – Joseph O'Rourke
    Commented Nov 19 at 16:32
  • $\begingroup$ Might be impossible if all polygons must be convex. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 19 at 18:29
  • $\begingroup$ I think it's still possible but the polygons get increasingly acicular. $\endgroup$
    – Noam D. Elkies
    Commented Nov 19 at 18:50

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Here is a picture (and extra characters to make it 30).

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  • $\begingroup$ This should work for any sequence, not just 3, 4, 5, etc. Each polygon must be of order at least 4 but triangles can always be combined to form higher-order polygons. $\endgroup$
    – Noam D. Elkies
    Commented Nov 20 at 6:03
  • $\begingroup$ Yes, this is quite universal. I knew a modification of this construction to reach an arbitrarily large minimum degree of the underlying graph (1-skeleton). $\endgroup$
    – Martin Tancer
    Commented Nov 20 at 6:38

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