Suppose that we reside in the set of all concave polygons (that is, polygons which are non-convex and simple, simple means that the boundary of the polygon does not cross itself).

Let us denote that set as $C$

The question is:

Does there exist $n_0 \in \mathbb N \setminus \{1,2\}$ and some concave polygon $p \in C$ such that $p$ can be tiled with $n_0$ congruent connected parts which are not polygons?

  • $\begingroup$ It might help our understanding of your question if you told us why you have excluded $n_0 = 2$. $\endgroup$ – Rob Arthan Nov 13 '15 at 23:21
  • $\begingroup$ @RobArthan Hey Rob, it is because I have some vague feeling that it maybe could be done for $n_0=2$, but maybe I am wrong. $\endgroup$ – Farewell Nov 13 '15 at 23:25
  • $\begingroup$ Do you require polygons to have straight line segments for their edges? $\endgroup$ – Rob Arthan Nov 13 '15 at 23:27
  • $\begingroup$ @RobArthan Yes. $\endgroup$ – Farewell Nov 13 '15 at 23:28
  • $\begingroup$ And a finite number of them, of course. $\endgroup$ – Farewell Nov 13 '15 at 23:31

Take a regular $n$-gon with centre $O$ and vertices $V_1, \ldots, V_n$ and draw a curvilinear arc $R_1$ from $O$ to $V_1$ that lies in the interior of $\triangle O V_1 V_2$. Now rotate $R_1$ through $2\pi k/n$ for $k = 1, 2, \ldots n - 1$, giving arcs $R_2, \ldots R_n$. This gives you $n$ congruent "curvilinear triangles" that tile the $n$-gon.

This image of a camera iris gives the idea much better than the above words.

To exhibit a similar tiling for a non-convex polygon, carry out the above construction for the polygon comprising the union of two regular $n$-gons with $n > 4$ that have a common edge (like two adjacent cells in a honeycomb). The boundary of the union is a non-convex simple polygon that can be tiled by the congruent non-polygons described above,

  • $\begingroup$ But those triangles overlap in a sense that they all have centre $C$ as they common point, and tiling is by definition composed of non-overlapping pieces, is it? $\endgroup$ – Farewell Nov 13 '15 at 23:46
  • $\begingroup$ No. The definition of a tiling requires that the tiles only intersect on their boundaries (see en.wikipedia.org/wiki/Tessellation). $\endgroup$ – Rob Arthan Nov 13 '15 at 23:51
  • $\begingroup$ Okay, but still regular n-gon is convex and the question is about concave polygons? $\endgroup$ – Farewell Nov 13 '15 at 23:53
  • $\begingroup$ See the revised answer. $\endgroup$ – Rob Arthan Nov 13 '15 at 23:59
  • $\begingroup$ If we take a non-convex polygon which is union of two regular n-gons that share an edge, then is that non-convex polygon simple? It seems to me that it is not because it has a boundary that crosses itself? $\endgroup$ – Farewell Nov 14 '15 at 0:06

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