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I start by stating the problem, which is already hinted in the title of the question. I do believe it is a research-level question.

Let us fix a regular $n$-gon with area $1$. What is the smallest possible area of a regular $m$-gon, so that it can cover the $n$-gon.

Of course, cover means to contain as a set. Of course, we can ignore the ill-defined scenarios, and fix $n\geq 3$ and $m\geq 3$. There is a number of situations on which it is possible to give a explicit formula (using sines and cosines) in terms of $n$ and $m$ of the desired smallest area.

Although this problem has a quite simple statement, it is not trivial for example what the answer is for $n=5$ and $m=7$ or for $n=7$ and $m=5$. Notice that the question only asks for a covering, so that it is not required to the polygons to be concentric.

When $m=4$ I could get an answer for all even $n$, and I think I can also solve it for odd $n$. Also, the case on which $n\mid m$ is, I think, much easier since the intuitive configuration does work, and a formula can be given.

Any thoughts on this? Is there any hope to give a formula using sums and products or quotients of sines and cosines of ugly rational multiples of $\pi$ for arbitrary $n$ and $m$?

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  • $\begingroup$ Do you know if the best concentric covering of a pentagon by a heptagon is in fact not the optimal covering? I.e., there is a non-concentric covering that beats every concentric covering? $\endgroup$
    – Joseph O'Rourke
    Commented Mar 3, 2021 at 16:08
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    $\begingroup$ For any $n$-gon the distance from centre to any point at $\theta$ which is already tilted at an angle $x$, is $d_n(\theta,x,R_n)=R_n\cos(\frac{\pi}{n})\sec(\frac{\pi}{n}-\langle {\frac{n(\theta-x)}{2\pi}} \rangle \frac{2\pi}{n})$, here $\langle {.} \rangle$ stands for the distance from nearest integer. As the area only depends on $r_m$, we have to find the $x$ such that $d_m(\theta,x)>d_n(\theta, 0)$ always and $R_m$ be minimum. Though it doesn't give direct formula, it may help finding the minimum value of $R_m$. $\endgroup$
    – Alapan Das
    Commented Mar 3, 2021 at 16:18
  • $\begingroup$ As follows from what @Robert Israel said (as an answer), the optimal configuration for $n=5$ and $m=7$ is obtained with non-concentric polygons. $\endgroup$
    – Luis Ferroni
    Commented Mar 3, 2021 at 21:20
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    $\begingroup$ Related question: how many intersection points do the borders of two polygons in an optimal configuration have? $\endgroup$
    – Max Alekseyev
    Commented Mar 5, 2021 at 16:04

1 Answer 1

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Let $p_i = [x_i,y_i]$ be the vertices of the $n$-gon. To minimize the $m$-gon covering it, we minimize $r$ subject to the constraints

$$ \cos(\theta + 2 \pi j/m) (x_i - x) + \sin(\theta+2\pi j/m) (y_i - y) \le r $$ with respect to the variables $\theta, x, y, r$. Here $(x,y)$ represents the coordinates of the centre of the $m$-gon.

In the case $n=5$, $m=7$, with $x_i = \cos(2 \pi i/5)$, $y_i = \sin(2\pi i/5)$, using numerical methods in Maple I find the minimum is approximately $r = 0.987554526096907$ for $\theta = \pi$, $x = 0.0378380934711730$, $y = 0$. On the other hand, the concentric case would have $x=0$, $y=0$, and a minimum $r$ of approximately $0.995974293995239$, which is significantly greater, again for $\theta=\pi$.

EDIT: Here is a picture of the optimal solution. The centre of the pentagon is in blue, the centre of the heptagon in red.

enter image description here

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    $\begingroup$ Could you clarify: Is $r$ the heptagon radius when the pentagon has unit area? Or did you use unit radius for the pentagon? $\endgroup$
    – Joseph O'Rourke
    Commented Mar 3, 2021 at 22:38
  • $\begingroup$ I would like this answer better with fewer digits. $\endgroup$
    – user44143
    Commented Mar 3, 2021 at 23:16
  • $\begingroup$ @JosephO'Rourke I thought it was obvious from the formulas: pentagon circumradius is $1$, heptagon inradius is $r$. $\endgroup$
    – Robert Israel
    Commented Mar 4, 2021 at 3:54
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    $\begingroup$ So, it looks like two odd-gons will always share an axis of symmetry. $\endgroup$
    – Max Alekseyev
    Commented Mar 5, 2021 at 16:07
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    $\begingroup$ That could well be true, though I'd hate to generalize from a sample size of $1$. $\endgroup$
    – Robert Israel
    Commented Mar 8, 2021 at 23:35

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