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(!!) We can only apply the claim when the sum two angles $\gamma$ forms with each side are $<\pi$ (indeed, in the claim the angle $\alpha(t)$ varies $<\pi$). However if the sum of angles in the claim was $\pi$, we can extend the curve by segments without increasing the curvature, and if the sum of angles is $>\pi$, we can achieve arbitrarily low curvature using the same trick. From now on we will assume that these sums of angles are $<\pi$. This is always true for polygons in which the sum of any two consecutive exterior angles is $<\pi$ (this is not a very restrictive condition, as all exterior angles add up to $2\pi$), but it may not work with triangles for example.

(!!) We can only apply the claim when the sum of the two angles $\gamma$ forms with each side is $<\pi$ (indeed, in the claim the angle $\alpha(t)$ has a total variation $<\pi$). However if the sum of the angles in the claim was $\pi$, we can extend the curve by segments without increasing the curvature, and if it is $>\pi$, we can achieve arbitrarily low curvature using the same trick. From now on we will assume that these sums of angles are $<\pi$. This is always true for polygons in which the sum of any two consecutive exterior angles is $<\pi$ (this is not a very restrictive condition, as all exterior angles add up to $2\pi$), but it may not work with triangles for example.

Suppose that $l$ is a side of $\mathcal{P}$ with length $d$ and the convex curve $c$ formed by segments and arcs of circumference is like in the figure below, forming angles $\alpha\geq\beta$ (with $\alpha+\beta<\pi$) with the side $l$.

Then it's not difficult to check that the radius of curvature of the circumference arc is $d\frac{\sin(\beta)}{\sin(\alpha+\beta)}\cot(\frac{\alpha+\beta}{2})$. Note that for fixed $\alpha+\beta$, the minimum curvature is achieved for $\alpha=\beta$, and in the case $\alpha=\beta\geq0$, the curvature increases with $\alpha$.

Suppose that $l$ is a side of $\mathcal{P}$ with length $x$ and the convex curve $c$ formed by segments and arcs of circumference is like in the figure below, forming angles $\alpha\geq\beta$ (with $\alpha+\beta<\pi$) with the side $l$.

Then it's not difficult to check that the radius of curvature of the circumference arc is $x\frac{\sin(\beta)(1+\cos(\alpha+\beta))}{\sin(\alpha+\beta)^2}$. Note that for fixed $\alpha+\beta$, the minimum curvature is achieved for $\alpha=\beta$, and in the case $\alpha=\beta\geq0$, the curvature increases with $\alpha$.

As Matt F. says, his answer is not optimal, but the optimal solution comes from a similar construction using just arcs of circumference and segments. This answer gives a finite dimensional family of curves containing the optimal curve, thus reducing the problem to minimizing some functions (with explicit expressions) from $[0,1]^n$ to $\mathbb{R}$.

Fix $(a,b)\in\mathbb{R}^2\setminus\{0\}$ with $a,b\geq0$ and some unit vector $v=(v_1,v_2)\in\mathbb{S}^1$ with $v_2\geq0$, and let $\Gamma$ be the set of $C^1$ curves $\gamma:[0,T_\gamma]\to\mathbb{R}^2$ parametrized by arc length such that:

Now back to the question: suppose that we have a convex curve $\gamma$ passing through all the points of $X$ (the points of $X$ are the vertices of a polygon $\mathcal{P}$, and maybe some additional points in edges of $\mathcal{P}$). Let $x,y$ be points of $X$ that appear consecutively in $\gamma$, with $\gamma(a)=x$, $\gamma(b)=y$ for some $a,b$. We can find some curve $f:[a,b]\to\mathbb{R}^2$ going from $x$ to $y$, with $f'(a)=\gamma'(a)$, $f'(b)=\gamma'(b)$ and such that $f$ has less curvature that $\gamma$ in [a,b]. We can summarize this in the following result:

Suppose that $l$ is a side of $\mathcal{P}$ with length $d$ and the convex curve $c$ formed by segments and arcs of circumference is like in the figure below, forming angles $\alpha\geq\beta$ with the side $l$.

Claim: If $X$ is the set of vertices of a polygon with a circumscribed circle $c$, then $c$ achieves min-max curvature.

As Matt F. says, his answer is not optimal, but the optimal solution, for most polygons (see (!!) below) comes from a similar construction using just arcs of circumference and segments. This answer gives a finite dimensional family of curves containing the optimal curve, thus reducing the problem to minimizing some functions (with explicit expressions) from $[0,1]^n$ to $\mathbb{R}$.

Fix $(a,b)\in\mathbb{R}^2\setminus\{0\}$ with $a,b\geq0$ and some unit vector $v=(v_1,v_2)\in\mathbb{S}^1$ with $v_2>0$, and let $\Gamma$ be the set of $C^1$ curves $\gamma:[0,T_\gamma]\to\mathbb{R}^2$ parametrized by arc length such that:

Now back to the question: suppose that we have a convex curve $\gamma$ passing through all the points of $X$ (the points of $X$ are the vertices of a polygon $\mathcal{P}$, and maybe some additional points in edges of $\mathcal{P}$). Let $x,y$ be points of $X$ that appear consecutively in $\gamma$, with $\gamma(a)=x$, $\gamma(b)=y$ for some $a,b$. We can find some curve $f:[a,b]\to\mathbb{R}^2$ going from $x$ to $y$, with $f'(a)=\gamma'(a)$, $f'(b)=\gamma'(b)$ and such that $f$ has less curvature that $\gamma$ in [a,b].

(!!) We can only apply the claim when the sum two angles $\gamma$ forms with each side are $<\pi$ (indeed, in the claim the angle $\alpha(t)$ varies $<\pi$). However if the sum of angles in the claim was $\pi$, we can extend the curve by segments without increasing the curvature, and if the sum of angles is $>\pi$, we can achieve arbitrarily low curvature using the same trick. From now on we will assume that these sums of angles are $<\pi$. This is always true for polygons in which the sum of any two consecutive exterior angles is $<\pi$ (this is not a very restrictive condition, as all exterior angles add up to $2\pi$), but it may not work with triangles for example.

We can summarize this in the following result:

Suppose that $l$ is a side of $\mathcal{P}$ with length $d$ and the convex curve $c$ formed by segments and arcs of circumference is like in the figure below, forming angles $\alpha\geq\beta$ (with $\alpha+\beta<\pi$) with the side $l$.

Claim: If the points of $X$ are in a circumference and the condition in (!!) is satisfied, then $c$ achieves min-max curvature.