Seeking a counterexample or proof to claim: "all regular polygons that can divided into triangles via straight lines can tile"

Question

Inspired by reading up on Tessellation I'm wondering if there is a general proof or counterexample to the following claim:

Any regular polygon that can be divided into triangles via straight lines (i.e. without any change of angle within the shape¹) will tile in the Euclidean plane

My intuition tells me that the claim is true. But to formulate a formal proof presents a challenge for me, I presume it involves using mathematical induction but I could be wrong.

---

_{¹E.g. like this, but not like this}

Answer 1

I think your notion of using straight lines to divide a polygon into triangles is different from what most other people would mean by those words.

Here is a pentagon that has been divided into triangles by straight lines.

What you seem to want is a polygon that can be divided into triangles by straight lines that intersect the polygon at its vertices and that pass through the center of the polygon.

Among polygons with more than four sides, no regular polygon with an odd number of sides can be subdivided in this way. But every regular polygon with an even number of sides can be divided into triangles by straight lines passing through two vertices and the center of the polygon. For example, a regular octagon can be subdivided in this way.

But you cannot tile a plane with regular octagons.

Answer 2

First, it is easy to show that every regular polygon can be divided in triangles, just connect the vertices with the center of the circumcircle.

Claim A regular $n$ polygon can tile the plane if and only if $n=3,4,6$.

Proof: If $n=3,4,6$ a tiling is trivial to find.

If $n \neq 3,4,6$ just calculate the angles of the $n$-gon and check that they are not divisors of $360$. This means that we cannot tile around any of the vertices of the $n$-gon.