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Let a simplex $\overline{ABC}$ in the plane $\mathbb{R}^2$ be defined as a triangle with vertices $A,B,C\in\mathbb{R}^2$ i.e. the subset $\overline{ABC}:=\{P\in\mathbb{R}^2 : P=\lambda_AA+\lambda_BB+\lambda_CC,\ \lambda_A,\lambda_B,\lambda_C\in[0,1], \lambda_A+\lambda_B+\lambda_C = 1\}$. Define a polygonal tile as a finite union of simplices $\mathcal{P}=\bigcup_{i=1}^n \overline{A_iB_iC_i}$ such that for $i\neq j$ the intersection of $\overline{A_iB_iC_i}$ and $\overline{A_jB_jC_j}$ is either empty, reduced to one of the vertices or a common side segment. Furthermore, we impose $\overset{\circ}{\mathcal{P}}$ to be path connected. Then show that every finite union $\mathcal{P}=\bigcup_{i=1}^n \overline{A_iB_iC_i}$ of simplices for which $\overset{\circ}{\mathcal{P}}$ is path connected is indeed a polygonal tile.

The assertion is pretty intuitive and it is very simple to show it in a concrete case. However, it seems very difficult to formally show it. I tried to use an induction on $n$, but if we remove a simplex from the union, we can no longer assume $\overset{\circ}{\mathcal{P}}$ to be path connected. Is there a somewhat simple way to show this?

Redundant Aunt
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The key is to drop the assumption that $\overset{\circ}{\mathcal{P}}$ is path connected. Then the same proof as given in Will Jaggy's answer to this question shows that every finite union of triangles $P$ has an interior edge. Next, you argue inductively: Given $P$ with a union $E$ of interior edges (pairwise disjoint except for their endpoints; I will call such edges "essentially disjoint"), you use the same argument as given by Jaggy to find an extra interior edge essentially disjoint from the rest of the interior edges, unless $P-E$ is a disjoint union of triangles. In the latter case, you get the desired decomposition of $P$ as the union of distinct triangles which intersect only along edges or vertices or not at all.

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Moishe Kohan
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