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Given some polygon $P$ in two-dimensional Euclidean space, I want to define the radius of $P$ as the average of the radii of the smallest outer circle and the largest inner circle. An outer circle has all points of $P$ inside of it (touching points allowed) and conversely, an inner circle has them outside of it. Note that their centers do not have to coincide. I want to show (as I suspect) that $2\pi$ is a lower bound for the perimeter of the polygon $P$ divided by the radius of $P$, i.e. that circles are the shape with the smallest value of "$\pi$".

My idea: If I fix the smallest outer and largest inner circle of $P$, I know they can be described by using their touching points. Keeping those fixed and changing everything else, it's easy to see how to minimize the circumference: You simply connect the points defining the inner circle directly (or, whenever they would have to traverse the inner circle, along that circle) with those defining the outer circle. Now, for the last step, it suffices to show that in both cases 1) the outer circle is fixed and 2) the inner circle is fixed: the minimum ratio (perimeter divided by radius) occurs when inner and outer circles coincide. This is where I am struggling. Any hints are appreciated!

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    $\begingroup$ Is there a reason you believe this is true? For example, is it an assigned exercise, or just a conjecture you’ve made? This affects what people can do to help. $\endgroup$ Mar 16, 2022 at 18:15
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    $\begingroup$ Thanks for pointing that out! I have played around with it a bit and came to this conclusion. It is a conjecture I have made up, however. $\endgroup$
    – Lyde
    Mar 16, 2022 at 18:18
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    $\begingroup$ Is the circumference of the polygon also the average of that of the two circles, or is it the actual perimeter of the polygon or of its convex hull? $\endgroup$
    – Henry
    Mar 16, 2022 at 18:19
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    $\begingroup$ It is certainly true for regular polygons, following from $\tan x>x$ for $x\in(0,\pi/2).$ $\endgroup$ Mar 16, 2022 at 20:19
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    $\begingroup$ Maybe a more correct conjecture would be requiring that the centre lies inside the $P$. $\endgroup$
    – Gargantuar
    Mar 17, 2022 at 16:06

1 Answer 1

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This theorem is certainly wrong, as the following counter example shows.

We consider an isosceles triangle $P$ with base $2$. Let its perimeter be $p$ and let the radii of the circumscribed and inscribed circle be $R$ and $r$ respectively.

Let $h$ be the height of $P$. We have $$p = 2 + 2\sqrt{1^2 + h^2}.$$ By the intersecting chords theorem, we have $$h(2R-h) = 1\times1 \iff 2\pi R = \pi\frac{1+h^2}{h}.$$ For our counter example to work, it suffices to show that $$2p < 2\pi R,$$ since this would imply $$2p < 2\pi R + 2\pi r \iff p\frac{2}{R+r} < 2\pi.$$

$2p < 2R$ can be shown with a lot of computation (isolating the root and squaring, giving a polynomial with degree four) that this is true for $$h \lesssim 0.450621.$$

The intuition: As $h \to 0$, we have $p \to 4$ and $2\pi R \to \infty$.

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