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Let $R_1$ be regular $n$-sided polygon on the plane (square, pentagon, hexagon, etc).
Now from this position we start to rotate this polygon about its center of gravity obtaining figure $R_2$.

  • How to calculate the angle of rotation $\alpha$ for the case where common area of $R_1$ and $R_2$ i.e. area ($R_1 \cap R_2)$ will be minimal? (intuition tells what possible solution could be but how to prove it?)
  • Does some simple method exist for solution of this problem in general case? (preferably with the use of rotations matrices)
  • The procedure for $n$-odd and $n$-even could be the same or we should to differentiate between these two cases?

Additionally:

  • Could it be proven that the shape obtained for situation of minimal area is also a regular polygon ( $2n$ sided) as we see in the below picture of pentagon made by Joseph?
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The minimum is achieved at $\alpha = \frac{\pi}{n}$ and at minimum, $R_1 \cap R_2$ is a regular $2n$-gon.

Choose a coordinate system so that $R_1$ is centered at origin and one of its vertices lies on $x$-axis.
Let $\rho(\theta)$ be the function which allow us to parametrize $\partial R_1$ in following manner:

$$\mathbb{R} \ni \theta \quad\mapsto\quad (x,y) = (\sqrt{2\rho(\theta)}\cos\theta,\sqrt{2\rho(\theta)}\sin\theta) \in \partial R_1$$

In terms of $\rho(\theta)$, we have

$$f(\alpha) \stackrel{def}{=} \verb/Area/(R_1 \cap R_2) = \int_0^{2\pi} \min(\rho(\theta),\rho(\theta-\alpha)) d\theta$$

Since $R_1$ is a regular $n$-gon and one of its vertices lies on $x$-axis, $\rho(\theta)$ is even and periodic with period $\frac{2\pi}{n}$. In fact, it strictly decreases on $[0,\frac{\pi}{n}]$ and strictly increases on $[\frac{\pi}{n},\frac{2\pi}{n}]$.

As a result of these, $f(\alpha)$ is even and periodic with same period. To determine the minimum of $f(\alpha)$, we only need to study the case where $\alpha \in \left[0,\frac{\pi}{n}\right]$.

For $\alpha \in \left[0,\frac{\pi}{n}\right]$ and $\theta \in \left[0,\frac{2\pi}{n}\right]$, the curve $\rho(\theta)$ and $\rho(\theta - \alpha)$ intersect at $\frac{\alpha}{2}$ and $\frac{\alpha}{2} + \frac{\pi}{n}$.
This leads to $$\begin{align}f(\alpha) &= n\left[ \int_{\frac{\alpha}{2}}^{\frac{\alpha}{2}+\frac{\pi}{n}} \rho(\theta) d\theta + \left( \int_0^{\frac{\alpha}{2}} + \int_{\frac{\alpha}{2}+\frac{\pi}{n}}^{\frac{2\pi}{n}} \right)\rho(\theta-\alpha)d\theta \right] = 2n\int_{\frac{\alpha}{2}}^{\frac{\alpha}{2}+\frac{\pi}{n}} \rho(\theta) d\theta\\ \implies \frac{df(\alpha)}{d\alpha} &= n\left(\rho\left(\frac{\alpha}{2}+\frac{\pi}{n}\right) - \rho\left(\frac{\alpha}{2}\right)\right) \end{align} $$ At the minimum, we have $$\frac{df(\alpha)}{d\alpha} = 0 \implies \rho\left(\frac{\alpha}{2}\right) = \rho\left(\frac{\alpha}{2} + \frac{\pi}{n}\right) = \rho\left(\frac{\alpha}{2} - \frac{\pi}{n}\right) = \rho\left(\frac{\pi}{n} - \frac{\alpha}{2}\right) $$ But $\frac{\pi}{n} - \frac{\alpha}{2}$ also belongs to $[0,\frac{\pi}{n}]$ and $\rho(\theta)$ is strictly decreasing there, this means

$$\frac{\alpha}{2} = \frac{\pi}{n} - \frac{\alpha}{2}\quad\implies\quad \alpha = \frac{\pi}{n}$$

Please note that this argument doesn't use the explicit form of regular $n$-gon. It uses

  • $n$-fold rotation symmetry about center,
  • $2$-fold reflection symmetry about a ray through a vertex,
  • $\rho(\theta)$ is strictly decreasing on suitable intervals of $\theta$.

This means the same argument should work for other shapes with similar properties. e.g. those obtain from filling the "interior" of a regular star polygon.

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  • $\begingroup$ what a fine solution .... it should be a classic.. and I like its general interpretation very much.. $\endgroup$ – Widawensen Nov 28 '16 at 10:53
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"Does some simple method exist for solution of this problem in general case?"

Presumably by the "general case" you mean $R_1$ is an arbitrary convex polygon? Or maybe an arbitrary simple polygon, perhaps nonconvex? I don't think this will have a simple answer. Below I computed that the minimum intersection area for the blue quadrilateral is achieved with $R_2$ rotated about $94.5^\circ$ degrees. In general you might have to resort to numerical optimization.


            MinTrap
And, Yes, the minimum for a regular pentagon is achieved at a rotation of $36^\circ$, half the $72^\circ$ angle subtended by each edge from the centroid. Etc.
            MinPentagon


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  • $\begingroup$ In, the case of pentagon, intuition gives us answer as you have presented , I expected it, but how to prove it provided regularity of the shape - it's not clear for me. If the shapes are overlapping it is trivial that the common area is maximal but why exactly in the half of regular angle for pentagon is minimal it is not known to me, only intuition tells that since there is a symmetry it has to be true. But how to use this symmetry in the proof ? And thank you Joseph for drawings, they made question much more available.. $\endgroup$ – Widawensen Nov 27 '16 at 22:23
  • $\begingroup$ I mean regular polygon but considered by some method independently from the number of sides i.e. $n$ is just the parameter used in the method but we don't concentrate whether it is square or hexagon. I understand that for irregular shapes it would be hard to find a general method, probably for complicated shapes only experimental method would be appropriate.. but maybe if we had some kind of a simple description of the shape we could infer somehow also about common area of the original figure and rotated one. $\endgroup$ – Widawensen Nov 27 '16 at 22:26

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