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\section{Introduction}

A strip tiling tiles a strip between 2 right angles of some fixed distance. A finite set of finite tiles is considered here.

Let $StripT$ be the set of lengths of strips that can be tiled. $$\left({l_1,l_2}\right)\in StripT \Rightarrow l_1+l_2\in StripT$$

An angle tiling tiles a sector of some fixed angle.

Let $AngleT$ be the set of lengths of strips that can be tiled. $$\left[{\left({\theta_1,\theta_2}\right)\in AngleT \land \theta_1+\theta_2\leq 2\pi}\right] \Rightarrow \theta_1+\theta_2\in AngleT$$

\section{Bootstrapped Tilings}

Let the tuples of sets $\left({Lengths_1,Angles_1}\right)$ and $\left({Lengths_2,Angles_2}\right)$ exist such that $Lengths_1\cap Lengths_2$ and $Angles_1\cap Angles_2$ are both empty (showing disjoint sets).

Suppose the following statement is true: $$\left[{Lengths_1\subset StripT\land Angles_1\subset AngleT}\right] \Leftrightarrow \left[{Lengths_2\subset StripT\land Angles_2\subset AngleT}\right] $$ Then fractal tilings can be made to bootstrap tilings of these strip lengths and sector angles. I would like an example nontrivial bootstrapped tiling or proof that such bootstrapping is impossible.

Such a bootstrapped tiling would likely use strip lengths or sector angles not in the sets in the tuples. Example:

\section{Regular Polygon Tilings}

The simplest tilings use regular polygons. A regular polygon with $n>2$ sides and angle $\theta<\frac{pi}{2}-\frac{pi}{n}$ has edge length $2\text{ArcCosh}\left({\frac{\cos\left({\frac{\pi}{n}}\right)}{\sin\left({\frac{\theta}{2}}\right)}}\right)$.

Some of these tilings neither fill a sector nor a strip.

To tile a sector, set $\theta=\frac{\pi}{k},k\in\mathbb{Z}^+$. This gives $$\left\{{\frac{\pi}{k} \mid k\in\mathbb{Z}\land k>1}\right\}\subset AngleT$$ and via addition: $$\left\{{q2\pi \mid q\in\mathbb{Q} \land 0<q\leq 1}\right\}\subset AngleT$$

To tile a strip, set $\theta=\frac{\pi}{2k},k\in\mathbb{Z}^+$. This gives $$\left\{{2\text{ArcCosh}\left({ \frac{\cos\left({\frac{\pi}{n}}\right)} {\sin\left({\frac{\pi}{4k}}\right)}}\right) \mid \left({n,k}\right)\subset\mathbb{Z}\land k>1 \land n>2}\right\}\subset LengthT$$

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\section{Simple bootstrapping from known tilable angle}

Let $\theta\in AngleT\land k\in\mathbb{Z}^+$. Consider a pentagon with angles $\left({\pi-\theta,\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2}}\right)$. Start tiling a strip length $l_1$. Let the pentagon have sides $\left({l_1,l_2,l_3,l_4,l_5}\right)$. The angle $\pi-\theta$ touches the vertex touching the sides $l_4,l_5$. Let $l_3=k*l_1+l\land l\in StripT$. $l_1$ bootstraps as a fractal. Todo: express $l_1$ in terms of $\theta,l$.

It would be nice to see lengths and sides which simultaneously bootstrap off of each other rather than requiring a known tiling for the angle.

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