# Tiling the concave polygons with non-polygons

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?

• It might help our understanding of your question if you told us why you have excluded $n_0 = 2$. – Rob Arthan Nov 13 '15 at 23:21
• @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. – Farewell Nov 13 '15 at 23:25
• Do you require polygons to have straight line segments for their edges? – Rob Arthan Nov 13 '15 at 23:27
• @RobArthan Yes. – Farewell Nov 13 '15 at 23:28
• And a finite number of them, of course. – 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.
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,
• 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? – Farewell Nov 13 '15 at 23:46