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We all know that the solution to the isoperimetric problem in the Euclidean plane is the circle. Unfortunately, the circle does not tile the plane without adding at least 2 bowed in triangles. Because of the shape formed by adding the triangles I looked at prototiles formed by the function: $f(j):=\begin{cases} Sin[x-\frac{1}{4}(3+(-1)^{1+j})\pi]^j+\frac{1}{2}(3+(-1)^{1+j}) & \\ Sin^j[x] \end{cases},\frac{\pi}{2}\leq \theta\leq \frac{(4+(-1)^{1+j})\pi}{2}, j \in \mathbb{N}$

I would show the first 12 examples but can't embed pictures yet.

I wrote an essay for fun on why they are nice and have a small perimeter to area ratio for a given string of fixed length and indicated that the best answer was likely for j=1,2 without proof. But they are not smooth all the way around since they have two cusp points on the left and right ends. I was wondering if there is any previous research generalizing the isoperimetric theorem to tilings. Is my function for j=1,2 the best? Does a tiling that is smooth all the way around the centroid exist?

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    Sorry about that. Realized it while you were commenting.2017-01-23

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A bounded domain with smooth boundary cannot tile a plane. Indeed, suppose $\Omega$ is such a domain. It has finitely many neighbors $\Omega_1, \dots,\Omega_n$ in the tiling. Let $\Gamma_j = \partial\Omega_j\cap \partial \Omega $ be the common part of the boundary of $\Omega_j$ and $\Omega$. These sets are closed, and together they cover $\partial\Omega$. Since $\partial\Omega$ is connected, there is a point $z$ that belongs to the boundaries of $3$ or more domains. It is impossible for all of these domains to have smooth boundaries at $z$, because each domain smooth boundary takes up a sector of angular size $\pi-\epsilon$ in a small neighborhood of $z$. There is not enough room within the total angle of $2\pi$.

any previous research generalizing the isoperimetric theorem to tilings

Theorem 4 of The Honeycomb Conjecture by Thomas C. Hales is an inequality of isoperimetric type for which the optimal domain is a regular hexagon. The Honeycomb conjecture itself concerns a tiling with "smallest total boundary length" (properly interpreted), also an problem of isoperimetric type.