The extreme polygons for the self
Chebyshev radius of the boundary

For a given metric space $(X,d)$, we denote by $B(x,r)$ the closed ball with center $x$ and radius $r$. If $M$ is a nonempty bounded subset of a metric space $(X,d)$, then by $\operatorname{diam}(M)=\sup_{x,y\in M}d(x,y)$ we denote the diameter of $M$, and by

$$r(M):=\inf\{a\geq 0\,|\,\exists\,x\in X,\,M\subset B(x,a)\}=\inf_{x\in X}\sup_{y\in M}d(x,y)$$

we denote the Chebyshev radius of $M$. A point $x_{0}\in X$ for which $M\subset B(x_{0},r(M))$ is called a Chebyshev center of $M$. In general, a Chebyshev center of a set is not unique, and therefore we denote by $Z(M)$ the set of all Chebyshev centers of a bounded set $M$. The mapping $M\mapsto Z(M)$ is known as the Chebyshev-center map.

Given a nonempty bounded subset $M$ of $X$ and a nonempty set $Y\subset X$, the relative Chebyshev radius (of the set $M$ with respect to $Y$) is defined by the formula

$$r_{Y}(M):=\inf\{a\geq 0\,|\,\exists\,x\in Y,\,M\subset B(x,a)\}=\inf_{x\in Y}\sup_{y\in M}d(x,y).$$

In the case $Y=M$, we get the definition of the relative Chebyshev radius of $M$ with respect to $M$ itself  i.e. the self Chebyshev radius of the set $M$:

$$\delta(M):=r_{M}(M)=\inf\{a\geq 0\,|\,\exists\,x\in M,\,M\subset B(x,a)\}=\inf\limits_{x\in M}\,\sup\limits_{y\in M}\,d(x,y),$$

(1)

see, e.g., [1], [3, p. 119], and [4].

It is clear that this value depends only on the set $M$ and the restriction of the metric $d$ to this set, hence it is an intrinsic characteristic of the (bounded) metric space $(M,d|_{M\times M})$. It has also the following obvious geometric meaning for compact $M$: $\delta(M)$ is the smallest radius of a ball having its center in $M$ and covering $M$.

The study of extremal problems for convex curves in the Euclidean plane with a given self Chebyshev radius started with paper [20] by Rolf Walter. In particular, he conjectured that $L(\Gamma)\geq\pi\cdot\delta(\Gamma)$ for any closed convex curve $\Gamma$ in the Euclidean plane, where $L(\Gamma)$ is the length of $\Gamma$ and $d$ is the Euclidean metric. In [20], this conjecture is proved for the case when $\Gamma$ is a convex curve of class $C^{2}$ and all curvature centers of $\Gamma$ lie in the interior of $\Gamma$. It is also shown that the equality $L(\Gamma)=\pi\cdot\delta(\Gamma)$ in this case holds if and only if $\gamma$ is of constant width; for sets of constant width the reader is referred to the monograph [17].

It is also proved in [20] that all $C^{2}$-smooth convex curves have good approximations by polygonal chains in terms of the self Chebyshev radius (1). This observation leads to natural extremal problems for convex polygons. In particular, the following result was also proved in [20]. For each triangle $P$ in the Euclidean plane, one has $L(\Gamma)\geq 2\sqrt{3}\,\cdot\delta(\Gamma)$ with equality exactly for equilateral triangles, where $\Gamma$ is the boundary of $P$. The proof of this result was simplified by Balestro, Martini, Nikonorova, and one of the present authors in [8], where the authors determined the self Chebyshev radius for the boundary of an arbitrary triangle. Moreover, some related problems were considered in detail in [8]. In particular, the maximal possible perimeter for convex curves and boundaries of convex $n$-gons with a given self Chebyshev radius were found.

As the authors of [8] discovered somewhat later, the original problem from [20], which was to prove the inequality $L(\Gamma)\geq\pi\cdot\delta(\Gamma)$ for any closed convex curve $\Gamma$ in the Euclidean plane, had already been solved many years ago in paper [11] by K.J. Falconer (the results of that paper were formulated in other terms, without using the self Chebyshev radius). An exposition of the corresponding result can also be found in [17, Theorem 4.3.2]. It should be noted that the following stronger result is also proved in [11]. Instead of convexity, it requires the curve to be rectifiable and instead of the plane it is in $\mathbb{R}^{n}$. Let $\Gamma$ be a closed rectifiable curve in $\mathbb{R}^{n}$ (with the Euclidean metric) such that for every point $x$ on $\Gamma$ there is a point of $\Gamma$ at distance at least $1$ from $x$. Then $\Gamma$ has length at least $\pi$, this value being attained if and only if $\Gamma$ bounds a plane convex set of constant width $1$.

This paper is devoted to the proof of Rolf Walter’s conjecture in [20] on quadrilaterals of minimum perimeter among all quadrilaterals with a given self Chebyshev radius of their boundaries that is as follows: $L(\Gamma)\geq\frac{4}{3}\sqrt{2\sqrt{3}+3}\,\cdot\delta(\Gamma)$ for any convex quadrilateral $P\subset\mathbb{R}^{2}$ with the boundary $\Gamma$.

Note that this inequality becomes an equality for quadrilaterals $P$ called magic kites (squares are not extremal in this sense). This definition is taken from [20] and means convex quadrilaterals which are hypothetically extreme with respect to the self Chebyshev radius. Up to similarity, such a quadrilateral could be represented by its vertices, that are as follows (see Fig. 1):

$$(-1,0),\quad(1,0),\quad\left(0,\frac{\sqrt{3}}{3}\sqrt{2\sqrt{3}+3}\right),\quad\left(0,-\frac{1}{3}\sqrt{2\sqrt{3}-3}\,\right).$$

The heights drawn from all the vertices to the sides farthest to them have the same length ($H_{i}$, $i=1,2,3,4$, denotes the bases of the corresponding heights on the sides of the considered quadrilateral), which coincides with the self Chebyshev radius of the boundary $\Gamma$ of this magic kite. Note that two such heights emanate from vertex $A$, one such height emanates from vertices $B$ and $D$, while the heights from vertex $C$ are shorter than the proper Chebyshev radius of the boundary of this polygon.

Note also that we have $L(\Gamma_{\square})=\frac{8}{\sqrt{5}}\,\cdot\delta(\Gamma_{\square})$, where $\Gamma_{\square}$ is a boundary of a square, and $\frac{8}{\sqrt{5}}>\frac{4}{3}\sqrt{3+2\sqrt{3}}=3.389946...$.

Our main result is the following theorem.

If a quadrilateral $P$ is not convex, then one can easy to find a convex quadrilateral $P_{1}$ with the same perimeter and such that the self Chebyshev radius of the boundary of $P_{1}$ is not less than the self Chebyshev radius of the boundary of $P$. Indeed, if the convex hull of a quadrilateral $ABCD$ is the triangle $ABD$, we can consider the point $C_{1}$ that is symmetric to $C$ with respect the straight line $BD$, see Fig. 2. The quadrilateral $ABC_{1}D$ is convex and has the same perimeter as $ABCD$ has. For any point $M\in[B,C]\bigcup[C,D]$, denote by $M_{1}$ the point on $[B,C_{1}]\bigcup[C_{1},D]$ that is symmetric to $M$ with respect the straight line $BD$. If we take any point $L\in[A,B]\bigcup[A,D]$, then the distance from $L$ to $M_{1}$ is not less that the distance from $L$ to $M$. This observation implies that the self Chebyshev radius of the boundary of $ABCD$ is not greater than the self Chebyshev radius of the boundary of $ABC_{1}D$. Therefore, the above theorem must be proved only for convex quadrilaterals.

We use mainly standard methods of Euclidean geometry and analysis to prove this theorem. Sometimes we have to resort to the help of computer algebra systems. In what follows we repeatedly use some well known properties of convex polygons. As a source on the properties of polygons, one can advise, in particular, books [12, 14, 18]. There is also a special book on the properties of quadrilaterals: [2].

The proof of the main theorem is technically quite difficult. We have to consider many special cases, and in each case we apply different methods. Despite all our attempts to simplify the proofs, we have not succeeded in doing so. We would like to believe that someone will find a shorter and more conceptual reasoning. But while this is not known, we decided to share our version of the proof.

For symbolic and numeric calculations, we used the well-established Maple, a general purpose computer algebra system (CAS). Wherever possible, we have used symbolic calculations. The Maple program provides a wide range of commands for working with symbolic expressions. Numerical calculations were used only where it was not possible to use symbolic calculations effectively and where there was no doubt about the correctness of the approximations (because there is some margin of magnitude with which to compare the result). The precision used was controlled by the Digits global variable, which has a default value of 10. However, it can be set to almost any other value. For more detailed information, see e. g. [16].

The paper is organized as follows. In Section 2 we consider some important information on Chebyshev radii and centers for bounded convex sets in an arbitrary Hilbert space. In Section 3 we prove some important results on self Chebyshev radii and centers for boundaries of convex polygons on the Euclidean plane. In Section 4 we describe a general plan of the study and obtain some auxiliary results. Sections 5, 6, 7, and 8 are devoted to the study of important partial cases of the general problem. We prove our main Theorem 1 on the basis of the results obtained in these sections. In the final section we consider some additional conjectures and outline the prospects for further research on the topic. The list of some important notations and definitions is given at the end of the paper.

The authors are grateful to Endre Makai, Jr., Horst Martini, and Yusuke Sakane for useful discussions and valuable advices. We are also grateful for the advices of both of the anonymous reviewers, whose comments and suggestions allowed us to improve the presentation of this article.

All propositions in this section can be found in various sources. We give them here in the form we need and provide brief proofs for the sake of completeness.

Let us consider any Hilbert space $H$ with the inner product $(\cdot,\cdot)$ and the corresponding norm $\|\cdot\|$ (in particular any $\mathbb{R}^{n}$ with some Euclidean norm). For any set $M\subset H$, the symbols $\operatorname{co}(M)$, $\operatorname{bd}(M)$, and $\operatorname{int}(M)$ denote respectively the convex hull, the boundary, and the interior of $M$.

For given $c\in H$ and $r\geq 0$, we consider $S(c,r)=\{x\in H\,|\,\|x-c\|=r\}$, $U(c,r)=\{x\in H\,|\,\|x-c\|<r\}$, and $B(c,r)=\{x\in H\,|\,\|x-c\|\leq r\}$, respectively the sphere, the open ball, and the closed ball with the center $c$ and radius $r$. For $x,y\in H$, the symbol $[x,y]$ means the closed interval with the ends $x$ and $y$ on the line through $x$ and $y$.

Proof. Let us consider any $x_{1},x_{2}\in H$ and any $\alpha,\beta>0$, $\alpha+\beta=1$. Then for every $z\in M$, we have

$$\|z-(\alpha x_{1}+\beta x_{2})\|=\|\alpha(z-x_{1})+\beta(z-x_{2})\|\leq\alpha\|z-x_{1}\|+\beta\|z-x_{2}\|\leq\alpha t(x_{1})+\beta t(x_{2}).$$

Therefore, $t(\alpha x_{1}+\beta x_{2})\leq\alpha t(x_{1})+\beta t(x_{2})$, hence, the function $t$ is convex. Since $M$ is bounded, then there is a real number $C>0$ such that $\|z\|\leq C$ for all $z\in M$. Consequently,

$$\|x\|-C\leq\|x\|-\|z\|\leq\|x-z\|\leq\|x\|+\|z\|\leq\|x\|+C$$

for every $x\in H$ and every $z\in M$. This means that $t$ is coercive, hence, $t$ has a point of global minimum value, see e. g. Corollary 11.30 in [9].

Now, we suppose that there are points $x_{2}\neq x_{1}$, where $t$ attains its minimum value, say $m_{t}$. Therefore, $M\subset B(x_{1},m_{t})$, $M\subset B(x_{2},m_{t})$, and $M\subset B(x_{1},m_{t})\cap B(x_{2},m_{t})$. It is clear that $B(x_{1},m_{t})\cap B(x_{2},m_{t})\subset B\left(\frac{x_{1}+x_{2}}{2},\sqrt{m_{t}^{2}-\|x_{2}-x_{1}\|^{2}/4}\right)$. Indeed, for any $z\in B(x_{1},m_{t})\cap B(x_{2},m_{t})$, we have

$$4\|z-(x_{1}+x_{2})/2\|^{2}+\|x_{2}-x_{1}\|^{2}=2\|z-x_{1}\|^{2}+2\|z-x_{2}\|^{2}\leq 4m_{t}^{2}.$$

Hence, $t\left(\frac{x_{1}+x_{2}}{2}\right)<m_{t}$, that is impossible.   

Proof. Let $z\in M$ be the closest point of $M$ to the point $y$. It is clear that $M\subset\{x\in H\,|\,\|x-z\|<\|x-y\|\}$. Hence if $M\subset B(y,r)$ for some $r>0$, then $M\subset B(z,r)$. It implies that $t(z)\leq t(y)$. If $H$ is finite-dimensional, then $M$ is compact, hence there is $u\in M$ such that $\|u-z\|\geq\|x-z\|$ for all $x\in M$. Therefore, $t(z)=\|u-z\|<\|u-y\|\leq t(y)$.   

For any non-empty bounded closed set $M$ in $H$, a point $y\in H$ ($y\in M$) with minimal values of the function $x\mapsto t(x)$ on $H$ (respectively, on $M$) is the Chebyshev center (respectively, the self Chebyshev center) for $M$. Two above propositions imply

Obviously $\operatorname{bd}(M)$ has the same Chebyshev center as $M$ from the above corollary. It is not the case for self Chebyshev centers, of course.

Proof. For any $\varepsilon>0$, there is a point $y_{\varepsilon}\in\operatorname{bd}(M_{2})$ such that $\operatorname{bd}(M_{2})\subset M_{2}\subset B(y_{\varepsilon},r_{2}+\varepsilon)$. Since $M_{1}\subset M_{2}$, we have $\operatorname{bd}(M_{1})\subset M_{1}\subset M_{2}\subset B(y_{\varepsilon},r_{2}+\varepsilon)$. If $y_{\varepsilon}\in\operatorname{bd}(M_{1})$, then we put $z_{\varepsilon}:=y_{\varepsilon}$. Now, suppose that $y_{\varepsilon}\not\in\operatorname{bd}(M_{1})$ and consider the function $t_{1}:H\rightarrow\mathbb{R}$ such that $t_{1}(x)=\min\{r\,|\,M_{1}\subset B(x,r)\}$. We see that $t_{1}(y_{\varepsilon})\leq r_{2}+\varepsilon$. By Proposition 2, there is $z_{\varepsilon}\in\operatorname{bd}(M_{1})$ such that $t_{1}(z_{\varepsilon})\leq t_{1}(y_{\varepsilon})$. In both cases we have $t_{1}(z_{\varepsilon})\leq t_{1}(y_{\varepsilon})\leq r_{2}+\varepsilon$. Since $\varepsilon>0$ could be as small as we want, we get $r_{1}=\inf\{t_{1}(z)\,|\,z\in\operatorname{bd}(M_{2})\}\leq r_{2}$.   

In what follows, the symbol $\|A\|$ means the cardinality of a given set $A$. We identify the Euclidean plane with $\mathbb{R}^{2}$ supplied with the standard Euclidean metric $d$, where $d(x,y)=\sqrt{(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}}$.

We call $\Gamma$ a convex curve if it is the boundary of some convex compact set in the Euclidean plane $\mathbb{R}^{2}$. Important examples of convex curves are convex closed polygonal chains, i.e. boundaries of convex polygons. A polygon $P$ is called an $n$-gon if it has exactly $n$ vertices. For $n=2$ we get line segments. The perimeter $L(P)$ of any $2$-gon is defined as the double length of the line segment $P$. Such definition is justified, since it leads to the continuity of the perimeter as a functional on the set of convex polygons with respect to the Hausdorff distance.

Let $\Gamma\subset\mathbb{R}^{2}$ be a compact set (in particular, a convex curve). We define the function $\mu:\Gamma\rightarrow\mathbb{R}$ as follows:

$$\mu(x)=\max\limits_{y\in\Gamma}d(x,y).$$

(2)

For any polygon $P$ and any given point $x\in P$, $\max\limits_{y\in P}d(x,y)$ is achieved at a vertex of $P$ (see, e.g., Lemma 4.1 in [20]), i.e., $\mu(x)$ is equal to the maximal distance from $x$ to vertices of $P$. Note also that the diameter $\operatorname{diam}(P)=\max\limits_{x,y\in P}d(x,y)$ of a polygon $P$ always connects two vertices.

The following property (monotonicity of the perimeter) of convex curves is well known (see, e.g., [10, §7]).

The following simple result is very useful.

Let us consider $n\geq 4$ and let $\Gamma$ be a boundary of a convex $n$-gon $P$. We will show how to estimate $\delta(\Gamma)$. Let us denote $A_{i}$, $1\leq i\leq n$, consecutive vertices of a polygonal line $\Gamma$ (vertices of the polygon $P$), assuming that indices are taken $\!\!\!\mod n$. The direction of traversing the curve $\Gamma$ in the direction $A_{1},A_{2},...,$ is called positive, and the opposite direction to it is called negative.

For a given point $A\in\Gamma$, we consider the set

$${\mathcal{F}}(A)=\{B\in\Gamma\,\mid\,d(A,B)=\mu(A)\},$$

(3)

see (2). It is clear that ${\mathcal{F}}(A)$ contains only vertices of $\Gamma$.

For a given point $A\in\Gamma$, we denote by $T_{+}(A)$ (respectively, $T_{-}(A)$) the limit $\lim\limits_{B\to A}\frac{\overrightarrow{AB}}{d(A,B)}$, where the points $B\in\Gamma$ tend to $A$ via the negative (respectively, the positive) direction. These vectors have length $1$ and are called one-sided tangent vectors at the point $A$. If $A$ is an interior point of some segment $[A_{i},A_{i+1}]$, then $T_{+}(A)=-T_{-}(A)$.

For a given point $A\in[A_{i},A_{i+1}]\subset\Gamma$ we put $N(A):=[A_{i},A_{i+1}]$ if $A$ is in the interior of the segment $[A_{i},A_{i+1}]$ and $N(A):=[A_{i-1},A_{i}]\cup[A_{i},A_{i+1}]$ if $A=A_{i}$ for some $i\in\mathbb{Z}_{n}$, see Fig. 3.

Proof. Obviously, 1) implies 2). Let us prove $2)\Rightarrow 3)$. Suppose that 2) holds but 3) does not hold. Choose a vector $\vec{v}\in\{T_{+}(A),T_{-}(A)\}$, such that $(\vec{v},\overrightarrow{AB})>0$ for any $B\in{\mathcal{F}}(A)$, and consider the points $A(\varepsilon)=A+\varepsilon\vec{v}\in\Gamma$. Since $(\vec{v},\overrightarrow{AB})>0$, then $d(A(\varepsilon),B)<d(A,B)=\mu(A)$ for sufficiently small $\varepsilon>0$. If $C$ is any vertex of $P$ that is not in ${\mathcal{F}}(A)$, then $d(A,C)<\mu(A)$, hence, $d(A(\varepsilon),C)<\mu(A)$ for sufficiently small $\varepsilon>0$. Therefore, $d(A(\varepsilon),D)<\mu(A)$ for any vertex $D$ of $P$, that implies $\mu(A(\varepsilon))<\mu(A)$ for sufficiently small $\varepsilon>0$. Hence, we get the contradiction with the condition 2) and, therefore, 2) implies 3).

Now, let us prove $3)\Rightarrow 1)$. We suppose that 3) holds. Let us choose any vector $\vec{v}\in\{T_{+}(A),T_{-}(A)\}$ and consider the point $A(\varepsilon)=A+\varepsilon\vec{v}$ for any $\varepsilon>0$. By the condition 3), there is $B\in{\mathcal{F}}(A)\subset\Gamma$ such that $(\vec{v},\overrightarrow{AB})\leq 0$. It implies $d(A(\varepsilon),B)>d(A,B)=\mu(A)$. Hence, $\mu(A(\varepsilon))\geq d(A(\varepsilon),B)>\mu(A)$ for all $\varepsilon>0$. Since we can take both $T_{+}(A)$ and $T_{-}(A)$ as $\vec{v}$, we get that $A$ is the minimal value point of $\mu$ on the set $N(A)$, hence, the condition 1) holds.   

Using the equivalence of conditions 1) and 3) in the previous proposition, we obtain the following

For a given point $A\in\Gamma$ we have $d(A,A_{i})=d(A,A_{j})$ if and only if $A$ is on the perpendicular bisector of $A_{i}$ and $A_{j}$. Hence, we have only finite numbers of points $A\in\Gamma$ such that the set ${\mathcal{F}}(A)$ has more that one point. Proposition 6 implies

Proof. By Corollary 3 we see that the set ${\mathcal{F}}(A_{i})$ has at least two elements. Let us consider the set $J=\{j\in\mathbb{Z}_{n}\,|\,A_{j}\in F(A_{i})\}=\{j\in\mathbb{Z}_{n}\,|\,d(A_{i},A_{j})=\mu(A_{i})\}$.

Now, we suppose that for every $k,l\in J$ we have $\angle A_{k}A_{i}A_{l}<\pi/2$. Let us fix $j\in J$ such that $j=i-1$ or $j=i+1$ (it is possible by the assumptions) and consider the points $A_{i}(\varepsilon)=\varepsilon A_{j}+(1-\varepsilon)A_{i}\in\Gamma$ for $\varepsilon\in(0,1)$. It is easy to see that

$$d(A_{i}(\varepsilon),A_{k})<d(A_{i},A_{k})=\mu(A_{i})$$

for all $k\in J$ and for sufficiently small $\varepsilon>0$ (due to the inequality $\angle A_{j}A_{i}A_{k}<\pi/2$). Moreover, for all $k\in\mathbb{Z}_{n}\setminus J$, $k\neq i$, we have $d(A_{i}(\varepsilon),A_{k})<\mu(A_{i})$ for sufficiently small $\varepsilon>0$ (due to the inequality $d(A_{i},A_{k})<\mu(A_{i})$). So we have $\mu(A_{i}(\varepsilon))<\mu(A_{i})$ for all sufficiently small $\varepsilon>0$, that is contradicts to $A_{i}\in G_{\operatorname{locmin}}$.

Therefore, there are $k,l\in J$ such that $\angle A_{k}A_{i}A_{l}\geq\pi/2$. It means that $P$ contains the triangle $\triangle A_{k}A_{i}A_{l}$, which has the perimeter at least $(2+\sqrt{2})\cdot\delta(\Gamma)$ (we have this minimal value for $\angle A_{k}A_{i}A_{l}=\pi/2$). Hence, $L(\Gamma)\geq(2+\sqrt{2})\cdot\delta(\Gamma)$.   

In what follows, we call an $n$-gon $P$ extremal if it has the smallest perimeter among all convex $n$-gons whose boundaries have a given value of the self Chebyshev radius (1) (i. e. $L(\Gamma)/\delta(\Gamma)$ has a minimal possible value, where $\Gamma=\operatorname{bd}(P)$).

The self Chebyshev radius $\delta(\Gamma)$ of $\Gamma$ coincides with the minimal value of the function $\mu$ (see (2)).

For fixed $n$-gon $P$ with $\Gamma=\operatorname{bd}(P)$, we consider the function $t:\mathbb{R}^{2}\rightarrow\mathbb{R}$ such that

$$t(x)=\min\{r\,|\,\Gamma\subset B(x,r)\}.$$

(4)

We know that this function is convex with an unique point of the global minimum $x_{0}$ (Proposition 1). The point $x_{0}$ is the Chebyshev center of $\Gamma$ (and $P$), $t(x_{0})$ is the Chebyshev radius $r_{\Gamma}$ of $\Gamma$ (and $P$). We know also that $x_{0}\in P$ by Corollary 1.

Since the function $t$ is convex, for any $r>r_{\Gamma}$, the set $U_{r}:=\{x\in\mathbb{R}^{2}\,|\,t(x)\leq r\}$ is a convex set, and $L_{r}:=\{x\in\mathbb{R}^{2}\,|\,t(x)=r\}$ is a closed convex curve.

Proof. If $x\in L_{r}$, then $\Gamma\subset B(x,r)$ and there is a point $z\in\Gamma$ such that $d(x,z)=r$. Obviously, $z$ is a vertex of $\Gamma$ (and $P$). Since we have only finite number of the vertices, we get the lemma.   

It is clear that

$$\delta(\Gamma)=\min\{r\in[r_{\Gamma},\infty)\,|\,\Gamma\cap L_{r}\neq\emptyset\}.$$

(5)

Hence, we have $U_{\delta(\Gamma)}\subset P$ and

$$G_{\min}:=L_{\delta(\Gamma)}\cap\Gamma=U_{\delta(\Gamma)}\cap\Gamma\neq\emptyset.$$

(6)

If $z\in G_{\min}$ and $z$ is a smooth point of $L_{\delta(\Gamma)}$, then $z$ is necessarily in the interior of some segment $[A_{i},A_{i+1}]\subset\Gamma$. If $z\in G_{\min}$ and $z$ is a non-smooth point (an intersection point of two circular arcs with different centers) of $L_{\delta(\Gamma)}$, then $z$ could be some vertex $A_{i}$ (in this case $\angle A_{i-1}A_{i}A_{i+1}\geq\pi/2$ by Corollary 2).

It is clear that the set $G_{\min}\cap[A_{i},A_{i+1}]$ is either empty or has exactly one point for any line segment $[A_{i},A_{i+1}]\subset\Gamma$.

Let us suppose that $P$ is extremal. Fix the vertex $A_{i}$ and consider two variations of $\Gamma$: $\Gamma_{\varepsilon}^{+}$ and $\Gamma_{\varepsilon}^{-}$, where $\varepsilon\in(0,1)$. This polygonal lines have the same vertices $A_{j}$ as $\Gamma$ has with one exception: instead of $A_{i}$ we take respectively $A_{i}^{+}({\varepsilon})=\varepsilon A_{i+1}+(1-\varepsilon)A_{i}$ and $A_{i}^{-}({\varepsilon})=\varepsilon A_{i-1}+(1-\varepsilon)A_{i}$. Since $A_{i}^{+}({\varepsilon})\in[A_{i},A_{i+1}]$ and $A_{i}^{-}({\varepsilon})\in[A_{i-1},A_{i}]$, then $L(\Gamma_{\varepsilon}^{+})<L(\Gamma)$ and $L(\Gamma_{\varepsilon}^{-})<L(\Gamma)$ for all $\varepsilon\in(0,1)$. Since $P$ is extremal, then $\delta(\Gamma_{\varepsilon}^{+})<\delta(\Gamma)$ and $\delta(\Gamma_{\varepsilon}^{-})<\delta(\Gamma)$ (by Proposition 3 we know that $\delta(\Gamma_{\varepsilon}^{+})\leq\delta(\Gamma)$ and $\delta(\Gamma_{\varepsilon}^{-})\leq\delta(\Gamma)$).

Now, we deal with the variation $\Gamma_{\varepsilon}^{-}$. There is a point $z_{\varepsilon}\in\Gamma_{\varepsilon}^{-}$ such that $d(z_{\varepsilon},x)\leq\delta(\Gamma_{\varepsilon}^{-})$ for any $x\in\Gamma_{\varepsilon}^{-}$, in particular, for all vertices $A_{j}$, $j\neq i$. Moreover, $d(z_{\varepsilon},A_{i}^{-}({\varepsilon}))\leq\delta(\Gamma_{\varepsilon}^{-})$. If we can choose $\varepsilon$ arbitrarily close to zero and such that $z_{\varepsilon}\in[A_{i}^{-}({\varepsilon}),A_{i+1}]$, then we get that $[A_{i},A_{i+1}]\cap G_{\min}\neq\emptyset$.

If $z_{\varepsilon}\not\in[A_{i}^{-}({\varepsilon}),A_{i+1}]$, then $z_{\varepsilon}\in\Gamma$, hence, $d(z_{\varepsilon},A_{i})>\delta(\Gamma_{\varepsilon}^{-})$ (otherwise $\delta(\Gamma_{\varepsilon}^{-})\geq\delta(\Gamma)$). If we can choose $\varepsilon$ arbitrarily close to zero and such that $z_{\varepsilon}\in\Gamma$, then we get some $z\in G_{\min}$ such that $d(z,A_{i})=\delta(\Gamma)$. Indeed, if $z_{\varepsilon}\to z$ as $\varepsilon\to 0$, then $d(z_{\varepsilon},A_{i}^{-}({\varepsilon}))\leq\delta(\Gamma_{\varepsilon}^{-})<d(z_{\varepsilon},A_{i})$, $A_{i}^{-}({\varepsilon})\to A_{i}$ and $\delta(\Gamma_{\varepsilon}^{-})\to\delta(\Gamma)$ as $\varepsilon\to 0$.

The variation $\Gamma_{\varepsilon}^{+}$ could be considered similarly. Hence we get the following

The main idea is to determine all extremal $4$-gons $P$ and to prove that this set coincides with magic kites. We use the notations from the previous sections.

Since $2+\sqrt{2}=3.414213...>\frac{4}{3}\sqrt{2\sqrt{3}+3}=3.389946...$, then we get the following simple consequence of Corollary 3 and Proposition 7.

Proof. By Corollary 4 we see that $A\not\in\{A_{i},A_{i+1}\}$. Suppose that $A_{i}\in{\mathcal{F}}(A)$, i. e. $d(A,A_{i})=\delta(\Gamma)$. By Proposition 6, for the vector $\vec{v}=\frac{1}{\|\overrightarrow{AA_{i}}\|}\overrightarrow{AA_{i}}\in\{T_{+}(A),T_{-}(A)\}$, there is a point $B\in{\mathcal{F}}(A)\subset\Gamma$ such that $(\vec{v},\overrightarrow{AB})\leq 0$. Hence, $\angle A_{i}AB\geq\pi/2$ and $d(A,A_{i})=d(A,B)=\delta(\Gamma)$. Since the triangle $A_{i}AB$ is a subset of the quadrilateral $P$ and has the perimeter $\geq(2+\sqrt{2})\cdot\delta(\Gamma)$, then $L(\Gamma)\geq(2+\sqrt{2})\cdot\delta(\Gamma)$. This means that $P$ is not extremal. Therefore, $A_{i}\not\in{\mathcal{F}}(A)$. Similar arguments proves $A_{i+1}\not\in{\mathcal{F}}(A)$.