An algorithm to find maximum area polygons circumscribed about a convex polygon

TL;DR

This paper introduces an algorithm to find the maximum area convex polygon circumscribed about a given convex polygon, providing solutions for general and regular polygons, and disproving a related conjecture.

Contribution

It presents a novel O(n^3) algorithm for maximum area circumscribed polygons and offers explicit solutions for regular polygons, also disproving a conjecture.

Findings

Algorithm computes maximum area circumscribed polygons in O(n^3) time

Explicit solutions provided for regular polygons

Disproves a conjecture of Farris

Abstract

A convex polygon Q is circumscribed about a convex polygon P if every vertex of P lies on at least one side of Q. We present an algorithm for finding a maximum area convex polygon circumscribed about any given convex n-gon in O(n^3) time. As an application, we disprove a conjecture of Farris. Moreover, for the special case of regular n-gons we find an explicit solution.

Equations

$$A(P)=\sup \{\mathrm{area}(Q):Qcircumscribed\; aboutP\},$$

$$\mathrm{area}(C)=\frac{1}{2}\sum _{k=1}^n|q_{k-1},q_k|=\sum _{k=1}^n|q_{k-1},p_k|=\sum _{k=1}^n\left|2\sum _{j=1}^{k-1}(-1{)}^{k-1-j}{p}_j+(-1{)}^{k-1}q,p_k\right|.$$

$$\mathrm{area}(C)=f(p_1,p_2,\dots ,p_n)+\sum _{k=1}^n|(-1{)}^{k-1}q,p_k|=f(p_1,p_2,\dots ,p_n)-\left|q,\sum _{k=1}^n(-1{)}^k{p}_k\right|,$$

$$\mathcal{G}(L)=2{\int }_0^1\left(p-L(p)\right)dp$$

Full text

1. Introduction

The algorithmic aspects of finding convex polygons under geometric constraints with some extremal property have been studied for a long time. We list just a few examples. Boyce et al. [7] dealt with the problem of finding maximum area or perimeter convex $k$-gons with vertices in a given set of $n$ points in the plane. Eppstein et al. [9] presented an algorithm that finds minimum area convex $k$-gons with vertices in a given set of $n$ points in the plane. Minimum area triangles [12, 15, 6] or more generally, convex $k$-gons [8, 3], enclosing a convex $n$-gon with $k<n$ were studied in several papers. Other variants, where area is replaced by another geometric quantity, were also investigated, see e.g. [13]. Maximum area convex polygons in a given simple polygon were examined, e.g. in [5, 14]. Algorithms to find polygons with a minimal number of vertices, nested between two given convex polygons, were presented in [6]. The authors of [16] examined among other questions the problem of placing the largest homothetic copy of a convex polygon in another convex polygon. For more information on geometry-related algorithmic questions, see [1].

Definition 1**.**

Let $P\subset\mathbb{R}^{2}$ be a convex $n$-gon. If $Q$ is a convex $m$-gon that contains $P$ and each vertex of $P$ is on the boundary of $Q$, then we say that $Q$ is circumscribed about $P$. Set

[TABLE]

if it exists. If $\operatorname{area}(Q)=A(P)$ and $Q$ is circumscribed about $P$, then $Q$ is a maximum area polygon circumscribed about $P$.

Note that $A(P)$ exists if and only if the sum of any two consecutive angles of $P$ is greater than $\pi$. In particular, this means that $P$ has at least five vertices. Furthermore, observe that if $Q$ is a maximum area polygon circumscribed about a convex polygon $P$, then every side of $Q$ contains at least one vertex of $P$. In the following we assume that these properties hold.

Given any convex polygon $P$, our aim is to find a maximum area convex polygon circumscribed about $P$. We investigate the properties of these polygons and present an algorithm to find them. Our results can be used to bound an integral of a positive convex function. As an application, we bound an integral of the Lorenz curve and disprove a conjecture of Farris about the Gini index in statistics.

This paper is organized as follows: In Section 2 we establish some geometric properties of maximum area polygons circumscribed about a convex $n$-gon. In Section 3 we present an algorithm with $O(n^{3})$ running time that finds $A(P)$ and the maximum area polygons circumscribed about $P$. Suppose that $Q$ is circumscribed about $P$. Let $S_{1},\ldots,S_{n}$ be sides of $P$ in counterclockwise order. We say that $S_{i}$ is “used” by $Q$ if it is on the boundary of $Q$, and “not used” otherwise. We can assign a sequence from $\{U,N\}^{n}$ to $Q$ such that the $i$th term is $U$ if $S_{i}$ is used and $N$ otherwise. In Section 4 we investigate the following problem: which sequences can be assigned to a maximum area circumscribed polygon, for some $P$. We do not have a full characterization, only partial results. In Section 5 we describe an application of our method to statistics, and disprove a conjecture of Farris in [10]. Finally, in Section 6 we collect our additional remarks and propose some open problems.

Throughout this paper, $P\subset\mathbb{R}^{2}$ denotes a convex $n$-gon, $n\geq 5$, and the sum of any two consecutive angles of $P$ is greater than $\pi$. The vertices of $P$ are denoted by $p_{1},p_{2},\ldots,p_{n}$, in counterclockwise order. We extend the indices to all integers so that indices are understood mod $n$; that is, we let $p_{i}=p_{j}$ if $i\equiv j\bmod n$. For any $i$, we denote the side $p_{i}p_{i+1}$ of $P$ by $S_{i}$. By $T_{i}$ we denote the triangle bounded by $S_{i}$ and the lines through $S_{i-1}$ and $S_{i+1}$; it is called the $i$th external triangle. Clearly, if $Q$ is circumscribed about $P$, then every vertex of $Q$ lies in the corresponding external triangle of $P$. Note that if three consecutive vertices of $Q$ lie on the same line, then removing the middle vertex does not change the area of $Q$. Thus, without loss of generality, in our investigation we deal only with circumscribed polygons without an angle equal to $\pi$. This implies, in particular, that if a side of $P$ is used (i.e. it is contained in the boundary of the circumscribed polygon $Q$), then it is contained in a side of $Q$.

2. Geometric properties of maximum area circumscribed

polygons

Theorem 1**.**

For any $i,j$ with $i\leq j\leq i+n$, let $Q$ be a convex polygon circumscribed about $P$ of maximal area containing $S_{j},S_{j+1},\ldots,S_{i+n}$ on its boundary: that is, these edges of $P$ are used by $Q$. Then for every $k=i+2,i+3,\ldots,j-1$ either

(a) $p_{k}$ is the midpoint of the side of $Q$ containing it,

or

(b) $S_{k-1}$ or $S_{k}$ lies on $\operatorname{bd}Q$.

Proof.

Assume that neither $S_{k-1}$ nor $S_{k}$ lies on $\operatorname{bd}Q$. This clearly implies that $p_{k}$ belongs to exactly one side of $Q$, we denote it by $V$. We show that in this case $p_{k}$ is the midpoint of $V$. Let $V_{+}$ (resp. $V_{-}$) be the side of $Q$ immediately after (resp. before) $V$ in the counterclockwise order, let $q_{+}=V\cap V_{+}$ and $q_{-}=V\cap V_{-}$, and let $L$, $L_{+}$, $L_{-}$ be lines through $V$, $V_{+}$ and $V_{-}$, respectively. Suppose that $p_{k}$ is not the midpoint of $V$. We can assume without the loss of generality that $|p_{k}q_{+}|>|p_{k}q_{-}|$. Rotate $L$ about $p_{k}$ by a very small angle $\alpha$ in the clockwise direction, denote the resulting line by $L^{\prime}$. Let $T_{+}$ be the triangle determined by $L_{+}$, $L$, and $L^{\prime}$, and let $T_{-}$ be the triangle determined by $L_{-}$, $L$, and $L^{\prime}$. Since $|p_{k}q_{+}|>|p_{k}q_{-}|$ and $\alpha$ is very small, $|p_{k}q_{+}|\alpha\approx\operatorname{area}(T_{+})>\operatorname{area}(T_{-})\approx|p_{k}q_{-}|\alpha$. Another possible argument is that since $|p_{k}q_{+}|>|p_{k}q_{-}|$ and $\alpha$ is very small, the reflection of $T_{+}$ about $p_{k}$ contains $T_{-}$. Thus modifying $Q$ by replacing $L$ by $L^{\prime}$ would increase its area. This implies that if $Q$ has the maximum area, then $p_{k}$ is the midpoint of $V$. ∎

Remark 1**.**

From the proof of Theorem 1 it is clear that it is valid for any maximum area polygon circumscribed about $P$.

Definition 2**.**

Let $1\leq k\leq n$ and let $q_{0},q_{1},\ldots,q_{k}$ be points in the plane. We say that the polygonal curve $C=q_{0}q_{1}\cdots q_{k}$ satisfies the midpoint property for the index $i$, if for $j=1,2,\ldots,k$, the vertex $p_{i+j}$ of $P$ is the midpoint of $q_{j-1}q_{j}$.

If $k=n$ and $q_{0}=q_{k}$, that is if $C$ is a closed polygonal curve and it satisfies the midpoint property for some index $i$, then we say that $C$ satisfies the midpoint property.

Theorem 2**.**

We have the following:

• (2.1)

If $n$ is odd, there is exactly one closed polygonal curve satisfying the midpoint property.

• (2.2)

If $n$ is even and $\sum_{k=1}^{n}(-1)^{k}p_{k}\neq 0$, then there is no closed polygonal curve satisfying the midpoint property.

• (2.3)

If $n$ is even and $\sum_{k=1}^{n}(-1)^{k}p_{k}=0$, then for every $q\in\mathbb{R}^{2}$, there is exactly one closed polygonal curve $C$ satisfying the midpoint property such that $q$ is the common endpoint of the two sides of $C$ containing $p_{1}$ and $p_{n}$. In addition, the absolute value of the signed area of $C$ is independent of $q$.

Proof.

Let $C=q_{0}q_{1}\cdots q_{n}$, $q_{0}=q_{n}$ be a closed polygonal curve satisfying the midpoint property for the index [math]. For every $k$, $q_{k}$ is the reflection of $q_{k-1}$ about $p_{k}$. Thus setting $q:=q_{0}$, we have $q_{1}=2p_{1}-q$, $q_{2}=2p_{2}-2p_{1}+q$, etc. In particular, since $C$ is closed, we obtain $q=2\sum_{k=1}^{n}(-1)^{n-k}p_{k}+(-1)^{n}q$. If $n$ is odd, it follows that $q=\sum_{k=1}^{n}(-1)^{n-k}p_{k}$, proving (2.1). If $n$ is even, it follows that $\sum_{k=1}^{n}(-1)^{n-k}p_{k}=\sum_{k=1}^{n}(-1)^{k}p_{k}=0$, implying (2.2) and the first part of (2.3).

Next we show that the signed area of $C$, also denoted by $\operatorname{area}(C)$, is independent of $q$. For any $u,v\in\mathbb{R}^{2}$, we denote by $|u,v|$ the determinant of the $2\times 2$ matrix with $u$ and $v$ as its columns. Since for every $k$, $|q_{k-1},q_{k}|=|q_{k-1},q_{k-1}+q_{k}|=2|q_{k-1},p_{k}|$, we obtain that

[TABLE]

Thus for some function $f=f(p_{1},p_{2},\ldots,p_{n})$, we have

[TABLE]

which is independent of $q$, since $\sum_{k=1}^{n}(-1)^{k}p_{k}=0$. ∎

Using the same idea, the following variant of Theorem 2 can be proved.

Theorem 3**.**

Let $1\leq k<n-1$. Let $\mathcal{C}$ denote the family of polygonal curves $C=q_{0}q_{1}\cdots q_{k}$ satisfying the midpoint property for the index $i$ such that $q_{0}$ lies on the line $L_{i-1}$ through $S_{i-1}$ and $q_{j}$ lies on the line $L_{i+j+1}$ through $S_{i+j+1}$.

• (3.1)

If $L_{i-1}$ and $L_{i+j+1}$ are not parallel, then $\mathcal{C}$ has exactly one element.

• (3.2)

If $L_{i-1}$ and $L_{i+j+1}$ are parallel and $L_{i+j+1}\neq 2\sum_{k=1}^{j}(-1)^{j-k}p_{i+k}+(-1)^{j}L_{i-1}$, then $\mathcal{C}=\emptyset$.

• (3.3)

If $L_{i-1}$ and $L_{i+j+1}$ are parallel and $L_{i+j+1}=2\sum_{k=1}^{j}(-1)^{j-k}p_{i+k}+(-1)^{j}L_{i-1}$, then for every $q_{0}\in L_{i-1}$ there is exactly one polygonal curve $C\in\mathcal{C}$ that starts at $q_{0}$. Furthermore, the signed area enclosed by $p_{i-1}q_{0}\cup C\cup q_{j},p_{i+j+2}\cup\left(\bigcup_{m=i+j+2}^{n+i-2}S_{m}\right)$ is independent of the choice of $q_{0}$.

Remark 2**.**

• •

The unique starting point $q$ in (2.1) can be found in $O(n)$-time. Same time is required for the computation of the solution $C$, for checking its convexity and for computing its area.

• •

In (2.3) the region for all possible starting points $q$, resulting in a convex solution, can be found in $O(n\log n)$ steps. Indeed, the polygonal curve satisfying the midpoint property is convex if any only if each vertex lies in the corresponding external triangle $T_{i}$ of $P$. Each of these conditions gives three linear constraints on the starting point $q$. The constraints can be obtained in $O(n)$ steps, and the intersection of these $3n$ halfplanes can be computed in $O(n\log n)$ time [4].

• •

If there exists a convex solution $Q$ in (2.3), then there is a convex solution $Q$ that contains a side of $P$. Indeed, if $q$ is on the boundary of the feasible region, then for some $k$, $q_{k}$ lies on a sideline of $P$, which yields that $Q$ contains a side of $P$.

• •

In (3.1) the unique starting point $q_{0}$, the corresponding solution $C=q_{0}q_{1}\ldots q_{j}$, its convexity properties and its area, can be found in $O(j)$ steps.

3. An algorithm to find the maximum area circumscribed polygons

For any $i,j$ with $i<j\leq i+n$, we define $Q_{ij}$ to be a maximum area convex polygon circumscribed about $P$ with the property that the sides $S_{j},S_{j+1},\ldots,S_{i+n}=S_{i}$ lie on the boundary of $Q_{ij}$. Let $A_{ij}=A_{ij}(P)$ be the area of some $Q_{ij}$. Note that in the case $j=i+n$, a polygon $Q_{i(i+n)}$ is restricted to contain the side $S_{i+n}=S_{i}$ in its boundary. We extend this definition to any ordered pair of integers $(i,j)$ by taking indices modulo $n$.

We present a recursive algorithm which computes $A_{ij}$ for all $i<j\leq i+n$. It also finds $A(P)$ and the maximum area circumscribed polygons about $P$.

It follows from the definition that for $j=i+1$, $Q_{ij}=P$. For $j=i+2$, we add an external triangle $T_{i+1}$ to $P$. Now let $2<k\leq n$ and suppose that we already know the value of $A_{i^{\prime}j^{\prime}}(P)$ for every $i^{\prime},j^{\prime}$ with $i^{\prime}<j^{\prime}<i^{\prime}+k$. Let $j=i+k$. Consider a polygon $Q$ circumscribed about $P$ such that the sides $S_{j},S_{j+1},\ldots,S_{i+n}$ lie on the boundary of $Q$. We distinguish between $k$ types of such polygons.

Type (0): $\operatorname{bd}Q$ does not contain any of the sides $S_{i+1},S_{i+2},\ldots,S_{j-1}$.

Type ($\alpha$): $\operatorname{bd}Q$ contains the side $S_{i+\alpha}$ for some $1\leq\alpha\leq k-1$.

Note that $Q$ can have several types, except for Type (0), which excludes the other types. We find the maximum area of a circumscribed convex polygon of each type separately.

Type (0): By Theorem 1, for any convex polygon $Q$ of maximum area, each of the vertices $p_{i+2},\ldots,p_{j-1}$ has to be the midpoint of the corresponding side of $Q$. Whether the sides $S_{i}$ and $S_{j}$ are parallel or no, the existence of $Q$ and its area can be found in $O(k)$-time. This follows from Theorem 3 and Remark 2.

Type ($\alpha$): By Theorem 1, an optimal solution $Q_{ij}$ is a union of some - by assumption already known - $Q_{i,i+\alpha}$ and $Q_{i+\alpha,j}$. It contains the sides $S_{j},S_{j+1},\ldots,S_{i+n}$ and $S_{i+\alpha}$, between $S_{i+n}$ and $S_{i+\alpha}$ it has the same vertices and sides as $Q_{i,i+\alpha}$, and between $S_{i+\alpha}$ and $S_{j}$ it has the same vertices and sides as $Q_{i+\alpha,j}$. Its area is $A_{ij}=A_{i,i+\alpha}+A_{i+\alpha,j}-\operatorname{area}(P)$. By construction, the convexity of $Q_{i,i+\alpha}$ and $Q_{i+\alpha,j}$ implies that $Q_{ij}$ is convex as well. Since $Q_{ij}$ can be of any Type ($\alpha$), this step requires $O(k)$-time.

For each fixed $k$, starting with $k=3$, we execute the above procedure for all $1\leq i\leq n$. Then we increase the value of $k$ by one and repeat all steps until $k=n$. We obtain the values of $A_{ij}(P)$ for all $i,j$ with $i<j\leq i+n$. This is done in $O(n^{3})$-time. Indeed, let $k$ be fixed, $3\leq k\leq n$. Only the case $j=i+k$ is unknown. $Q$ can be of any type, by above all types require $O(k)$-time. Executing this for all $i$, $1\leq i\leq n$, requires $O(kn)$-time. Now we have $A_{ij}$ for $i<j\leq i+k,1\leq i\leq n$ with $A_{is}\leq A_{il}$ for $s<l$. Hence it remains to take the maximum of $A_{i(i+k)}$ over $i$, which requires $O(n)$-time. Thus for a fixed $k$, the algorithm needs $O(kn)$-time. Summing over $k$, we obtain the claimed $O(n^{3})$-time.

Once we have $A_{i(i+n)}(P)$ for all $i$, we can calculate the maximum area $A_{\geq 1}(P)$ of a convex polygon circumscribed about $P$ containing at least one side of $P$. $A_{\geq 1}(P)=\max\{A_{i(i+n)}\ :\ 1\leq i\leq n\}$.

Denote by $A^{\prime}=A^{\prime}(P)$ the maximum area of a convex polygon circumscribed about $P$ containing none of the sides of $P$. By Remark 2, for even $n$, if there exists a convex solution containing none of the sides of $P$, then there is a solution containing one side of $P$ with the same area, hence $A^{\prime}\leq A_{\geq 1}$. However, if we would like to list all maximum area polygons circumscribed about $P$, we need to execute this step also in case $n$ is even. For an odd $n$, the existence of a convex solution, the solutions and their area can be computed in $O(n)$-time, see (2.1) and Remark 2. For an even $n$, all convex solutions can be found in $O(n\log n)$-time, see (2.3) and Remark 2.

Finally, $A(P)=\max\{A^{\prime}(P),A_{\geq 1}(P)\}$, so we get the final answer in $O(n^{3})$ time. It is clear that we can keep track of the best circumscribed polygons of different types throughout the algorithm. Hence, in addition to $A(P)$, we also get the polygons $Q$ realizing it.

4. Combinatorial properties of maximum area circumscribed polygons

Let $Q$ be a maximum area convex polygon circumscribed about $P$. Recall that a side of $P$ is called used, if it lies on $\operatorname{bd}Q$, and not used otherwise. Thus to any such $Q$, we associate an $n$-element sequence $s$ of $U$s and $N$s in such a way that the $i$th element of this sequence is $U$ if $S_{i}$ is used and $N$ otherwise. In particular, for each value of $i$, $s$ determines whether the condition (a) or (b) of Theorem 1 is satisfied for $p_{i}$. In this section we show that most $n$-element sequences are indeed assigned to some maximum area circumscribed polygon for a suitably chosen $P$. Our aim is to determine which sequences $s\in\{U,N\}^{n}$ are realizable; that is, which sequences can appear for some $P$ (and a maximum area convex polygon $Q$ circumscribed about $P$). We do not have a complete description, but we can tell which sequences can appear as subsequences.

Theorem 4**.**

(a) Let $s\in\{U,N\}^{k}$. It is a subsequence of a realizable sequence $s^{\prime}\in\{U,N\}^{n}$ for some $n\geq k$ if and only if $s$ does not contain three consecutive $U$s.

(b) Let $s\in\{U,N\}^{n}$ be a sequence that contains at least two nonconsecutive $U$s, and no three consecutive $U$s. If $s$ is composed of exactly two disjoint subsequences of $N$s separated by some $U$s, assume that one of the two subsequences has length at least $4$. Then $s$ is realizable.

Proof.

For any $x,y,z\in\mathbb{R}^{2}$, we denote the triangle $\operatorname{conv}\{x,y,z\}$ by $[x,y,z]$ and its area by $A(x,y,z)$.

First observe that no realizable sequence contains three consecutive $U$s. Indeed, in this case adding to $Q$ the triangle bounded by these three sidelines strictly increases the area of $Q$ while preserving its convexity. This proves one implication of (a). Second implication follows from part (b).

The rest of proof is based on the following two technical statements about realizing a long sequence of $N$s as a subsequence.

• (i)

For any $i\geq 1$ there exists a convex polygonal curve $\Gamma_{i}=p_{0}p_{1}\ldots p_{i}$ and two halflines, $L$ and $L^{\prime}$, starting at $p_{0}$ and $p_{i}$, respectively, such that the triangle $T_{i}$, bounded by $L$, $L^{\prime}$ and $p_{0},p_{i}$ contains $\Gamma$ and so that the following property is satisfied. Let $\Delta_{i}=q_{1}q_{2}\ldots q_{i}$ be a convex polygonal curve with $q_{1}\in L$, $q_{i}\in L^{\prime}$, and $p_{j}\in q_{j}q_{j+1}$ for $1\leq j\leq i-1$. Assume that the area of the convex polygon $p_{0}q_{1}q_{2}\ldots q_{i}p_{i}$ is maximal among all convex polygons with the above constraints. Then $q_{1}q_{2}\ldots q_{i}$ satisfies the midpoint property; that is, $p_{j}$ is the midpoint of $q_{j}q_{j+1}$ for $1\leq j\leq i-1$ (cf. Figure 1).

• (ii)

For any $i\geq 4$ there exists a convex polygonal curve $\Gamma_{i}=p_{0}p_{1}\ldots p_{i}$ and a point $p$ so that the polygon $pp_{0}p_{1}\ldots p_{i}$ is convex and so that the following property is satisfied.

Let $L$ be the half-line of the line through $pp_{0}$, which starts at $p_{0}$ and does not contain $p$, and let $L^{\prime}$ be the half-line of the line through $pp_{i}$, which starts at $p_{i}$ and does not contain $p$. Let $\Delta_{i}=q_{1}q_{2}\ldots q_{i}$ be a convex polygonal curve with $q_{1}\in L$, $q_{i}\in L^{\prime}$, and $p_{j}\in q_{j}q_{j+1}$ for $1\leq j\leq i-1$. Assume that the area of the convex polygon $pq_{1}q_{2}\ldots q_{i}$ is maximal among all convex polygons with the above constraints. Then $q_{1}q_{2}\ldots q_{i}$ satisfies the midpoint property; that is, $p_{j}$ is the midpoint of $q_{j}q_{j+1}$ for $1\leq j\leq i-1$ (cf. Figure 2).

First we show how part (b) of Theorem 4 follows from (i) and (ii). Let us decompose $s$ into $k$ consecutive subsequences $n_{1},n_{2},\ldots,n_{k}$ consisting only of $N$s that are separated by either $U$ or by $UU$.

If $k\geq 3$, we only need (i). Let $\bar{P}_{k}=r_{0}r_{1}\ldots r_{k}$, where $r_{0}=r_{k}$, be a regular $k$-gon. Sides $r_{i}r_{i+1}$ of $\bar{P}_{k}$ will be called old sides. Now we add one or two very small sides, called new sides at each vertex of $\bar{P}_{k}$, according to the number of $U$s separating $n_{m-1}$ and $n_{m}$. When we add one new side at $r_{i}$, we let it have the same angle with the two consecutive old sides. When we add two, we allow them to have almost the same angle with the two consecutive old sides. Let $\bar{P}$ denote the resulting polygon. Let $\bar{T}_{m}$ be the external triangle bounded by the $m$th old side and the lines of the adjacent new sides. If $n_{m}$ consists of $i$ $N$s, let $h_{m}$ be the affine transformation such that the image of the triangle $T_{i}$ in (i) is $h_{m}(T_{i})=\bar{T}_{m}$, and the image of $p_{0},p_{i}$ is the $m$th old side of $\bar{P}$. Then, replace this side of $\bar{P}$ by $h_{m}(\Gamma_{i})$ for all values of $m$. We obtain a convex polygon $P$. If the new sides are sufficiently small, then for any maximum area polygon circumscribed about $P$ these sides are used. On the other hand, from (i) it follows that no other sides of $P$ are used. If $k=2$ and say $n_{1}\geq 4$, then we start our construction with a segment and two nonparallel lines through its endpoints. On one side we apply (ii) with a suitable affine transformation and on the other side we apply (i) with a suitable affine transformation.

Next, we show that (i) holds. Since it is trivial if $i=1$ or $i=2$, we prove it first for $i=3$. Consider an isosceles triangle $[p_{1},p_{2},y_{2}]$ with base $p_{1}p_{2}$. Let $x_{2}$ be an arbitrary interior point of $p_{1}p_{2}$, and $q_{2}$ be an arbitrary point of $y_{2}x_{2}$, very close to $y_{2}$. Reflect the points $y_{2},q_{2}$ and $x_{2}$ about $p_{1}$ to obtain the points $p_{0},q_{1},y_{1}$, respectively. Reflect also the points $y_{2},q_{2}$ and $x_{2}$ about $p_{2}$ to obtain $p^{\prime}_{3}$, $q_{3}$ and $y^{\prime}_{3}$, respectively (see Figure 3). Then we clearly have $A(p_{1},p_{2},y_{2})=A(p_{0},p_{1},y_{1})+A(p_{2},p^{\prime}_{3},y^{\prime}_{3})=A(p_{0},p_{1},q_{1})+A(p_{1},p_{2},q_{2})+A(p_{2},p^{\prime}_{3},q_{3})$. Now slightly rotate the line of $p^{\prime}_{3}y^{\prime}_{3}$ about $q_{3}$ so that it intersects $p_{2}y^{\prime}_{3}$ at an interior point $y_{3}$. Let $p_{3}$ be the intersection point of this rotated line and the line of $p_{2}p^{\prime}_{3}$. By the idea of the proof of Theorem 1, we have $A(p_{0},y_{1},p_{1})+A(p_{2},p_{3},y_{3})<A(p_{1},p_{2},y_{2})<\sum_{j=1}^{3}A(p_{j-1},q_{j},p_{j})$. Furthermore, if $q_{2}$ is sufficiently close to $y_{2}$, then for any $q^{\prime}_{1}\in p_{0}y_{1}$ and $q^{\prime}_{2}\in y_{2}p_{2}$, if $p_{1}$ is the midpoint of $q^{\prime}_{1}q^{\prime}_{2}$, then $A(p_{0},q^{\prime}_{1},p_{1})+A(p_{1},q^{\prime}_{2},p_{2})<\sum_{j=1}^{3}A(p_{j-1},q_{j},p_{j})$, and a similar statement holds if we choose points $q^{\prime}_{2}\in p_{1}y_{2}$ and $q^{\prime}_{3}\in p_{3}y_{3}$ in the same way. Set $\Gamma_{3}=p_{0}p_{1}p_{2}p_{3}$, let $L$ be the half-line containing $p_{0}q_{1}$ and starting at $p_{0}$, and let $L^{\prime}$ be the half-line containing $p_{3}q_{3}$ and starting at $p_{3}$. Then the conditions of (i) are satisfied.

Now, we show how to modify this construction for larger values of $i$. We start with the configuration in the last paragraph. Let $\Delta_{3}$ be the curve that satisfies the conditions in (i). Rotate the line through $y_{3}p_{3}$ around $q_{3}$ by a very small angle such that the rotated line intersects $p_{2}p_{3}$ at an interior point $p^{\prime\prime}_{3}$. Let this line intersect the line through $p_{2}y_{3}$ at $y^{\prime\prime}_{3}$. Reflect the points $y^{\prime\prime}_{3},q_{3},p^{\prime\prime}_{3}$ about $p_{3}$ to obtain the points $p_{4},q_{4},y_{4}$, respectively (cf. Figure 4). Note that as the angle of rotation tends to zero, for the convex polygonal curve $\Delta_{4}=q_{1}q_{2}q_{3}q_{4}$ satisfying the conditions, the initial part $q_{1}q_{2}q_{3}$ will get arbitrarily close to $\Delta_{3}$. Since for a very small rotation angle $A(p_{3},q_{4},p_{4})>0$, $\Delta_{4}=q_{1}q_{2}q_{3}q_{4}$ does not use any of the sides, so it satisfies the conditions. We can proceed similarly and extend our construction for any $i$.

Now we modify our construction in (i) to show part (ii). First, let $i=4$. We define $p_{0}$, $p_{1}$, $p_{2}$, $p^{\prime}_{3}$, $q_{1}$, $q_{2}$, $q_{3}$, $y_{1}$, $y_{2}$ and $y_{3}$ as in Figure 3, but relabel $y_{2}$ as $y^{\prime}_{2}$ (see Figure 5).

We obtain $y_{2}$ by moving $y^{\prime}_{2}$ away from $p_{1}$ on the line through $p_{1}y^{\prime}_{2}$, and set $p_{3}$ as the intersection of the lines through $y_{2}p_{2}$ and $y^{\prime}_{3}p^{\prime}_{3}$. Note that in this case we have $A(p_{1},y_{2},p_{2})>A(p_{0},y_{1},p_{1})+A(p_{2},y^{\prime}_{3},p_{3})=\sum_{j=1}^{3}A(p_{j-1},q_{j},p_{j})$. Now, choose a point $y_{3}$ on the line through $p_{2}y^{\prime}_{3}$ such that $y^{\prime}_{3}$ is an interior point of $p_{2}y_{3}$, and $y_{3}$ is very close to $y^{\prime}_{3}$ (see Figure 6). Let $q_{4}$ be the reflection of $q_{3}$ about $p_{3}$. Consider a point $p_{4}$ on the line through $y_{3}p_{3}$ such that the distance of $p_{3}$ from $y_{3}$ is much smaller that from $p_{4}$. Finally, denote the intersection of the lines through $p_{2}p_{3}$ and $p_{4}q_{4}$ by $y_{4}$, and by $L$ and $L^{\prime}$ those half-lines of the lines through $p_{0}y_{1}$ and $p_{4}y_{4}$, starting at $p_{0}$ and $p_{4}$, that intersect and let $p$ be their intersection. Then the polygon $pp_{0}\ldots p_{4}$ is convex.

Observe that if $y_{3}$ is sufficiently close to $y^{\prime}_{3}$, and $p_{4}$ is sufficiently far from $p_{3}$ (i.e. $A(p_{3},y_{3},y^{\prime}_{3})$ is small and $A(p_{3},p_{4},q_{4})$ is large), then $A(p_{0},y_{1},p_{1})+A(p_{2},y_{3},p_{3})$ $<A(p_{1},y_{2},p_{2})+A(p_{3},y_{4},p_{4})$ $<\sum_{j=1}^{4}A(p_{j-1},q_{j},p_{j})$. It follows now directly that conditions in (ii) are satisfied by $\Gamma^{\prime}_{4}=p_{0}\ldots p_{4}$. To prove (ii) for $i>4$ we proceed exactly as in the proof of (i). ∎

Remark 3**.**

Let $P$ be a regular $n$-gon with unit circumradius, where $n\geq 5$. Then an elementary (but tedious) computation yields the following.

1. (1)

If $2|n$, then $A(P)=n\tan\frac{\pi}{n}+\frac{n}{2}\frac{\sin^{4}\frac{\pi}{n}}{\cos^{2}\frac{\pi}{n}}\tan\frac{2\pi}{n}$, and the sequence assigned to a maximum area polygon circumscribed about $P$ is $UNUN\ldots UN$. 2. (2)

If $4|n-1$, then $A(P)=n\tan\frac{\pi}{n}+\frac{\sin^{4}\frac{\pi}{n}}{\cos^{2}\frac{\pi}{n}}\left(\frac{n+3}{4}\tan\frac{2\pi}{n}-\tan\frac{3\pi}{2n}\right)$, and the sequence assigned to a maximum area polygon circumscribed about $P$ is $U\overbrace{NN\ldots N}^{\frac{n-5}{2}}UNUN\ldots UN$. 3. (3)

If $4|n+1$, then $A(P)=n\tan\frac{\pi}{n}+\frac{\sin^{4}\frac{\pi}{n}}{\cos^{2}\frac{\pi}{n}}\left(\frac{n+1}{4}\tan\frac{2\pi}{n}-\tan\frac{\pi}{2n}\right)$, and the sequence assigned to a maximum area polygon circumscribed about $P$ is $U\overbrace{NN\ldots N}^{\frac{n-3}{2}}UNUN\ldots UN$.

5. An application to statistics

The above content has a connection to the Gini index, a measure originally used in economics, statistics, and nowadays being used in many applications, see [10] for a very nice introduction to this subject.

In economics, the Lorenz curve is a representation of the distribution of wealth, income, or some other parameter. For a population of size $n$, with values (say, wealth) $x_{i}$ in increasing order, $F_{i}=i/n$, $S_{i}=\sum_{j=1}^{i}x_{j}$, and $L_{i}=S_{i}/S_{n}$. Then the function $L(F):$ $F_{i}\longrightarrow L_{i}$ is the Lorenz curve of the given distribution. That is, $L_{i}$ is the relative share of the poorest $i/n$ part of the population from the total wealth.

In general, for $0\leq\alpha\leq 1$, let $x_{\alpha}$ denote the $\alpha$-quantile of a distribution, that is, exactly $\alpha$ portion of the population has wealth less than $x_{\alpha}$. Let $\bar{x}$ be the mean of the distribution. Then the Lorenz curve is the function $L(p):=\tfrac{1}{\bar{x}}\int_{0}^{p}x_{\alpha}\ d\alpha$ for $0\leq p\leq 1$. Clearly, $L(0)=0$ and $L(1)=1$ and $L$ is always a convex function. $L(p)=p$ for every $p$ iff everybody has exactly the same wealth.

The functional

[TABLE]

is called the Gini coefficient. It measures the relative area between a neutral scenario and the observed scenario (See figure  7). (More precisely, twice the area between a neutral scenario and the observed scenario divided by the area under the curve for the neutral scenario.) It is [math] in case of “perfect equality” (everybody has the same wealth) and (almost) $1$ in case of “perfect inequality” (one person has all the wealth).

We defined the Lorenz curve as a continuous curve. In practice often only points on the Lorenz curve are known.

Remark 4**.**

Data for every individual is often not available and only data for groups is accessible. From that data only points of the Lorenz curve can be reconstructed.

Remark 5**.**

In credit modeling, banks group their clients in $n$ rating groups. After twelve months they see which clients could not pay back their loans and the Lorenz curve is then taken as the percentage of defaults in the worst $i$ groups. A high Gini coefficient indicates that the bank succeeded in discrimating safe clients from dangerous clients.

For both cases above, the real Gini coefficient is not known and upper and lower bounds are of interest. By the convexity of a Lorenz curve the best lower bound is attained by the polygonal curve obtained by connecting the known points on the Lorenz curve. On the other hand, whereas it is easy to see that any maximal area convex curve must be piecewise linear, it does not seem easy to find the best upper bound corresponding to any given point set. A conjecture related to this problem, made by Farris [10], states that the maximal value is attained at a convex polygonal curve with the property that each side of it lies on a sideline of the polygonal curve connecting the given points, or, using our terminology, no sequence associated to any polygonal curve contains consecutive $N$s.

Our results imply that Farris’ conjecture does not hold. As a specific counterexample, we may take the part of a regular $n$-gon $P_{n}$ with $8|n-4$, centered at $(0,1)$, with a vertex at $(0,0)$ and contained in the unit square $[0,1]^{2}$. From the computations proving Remark 3 it is easy to see that in this case the optimal circumscribed polygonal curve does not use any sides of $P_{n}$. Equivalently, using the idea of the proof of Theorem 4, we may construct counterexamples assigned to ‘almost all’ sequences of $U$s and $N$s.

Moreover, our algorithm from Section 3 provides an efficient way to find the best upper bound for the Gini index for any given points of the Lorenz curve.

6. Remarks and questions

Problem 1**.**

For any value of $n$, determine the $n$-element sequences of $U$s and $N$s that are realizable.

It is also natural to investigate the following higher-dimensional generalization of our problem.

Problem 2**.**

Given a convex polytope $P$ in a Euclidean $d$-space, find the maximum volume convex polytopes $Q$ with the property that each vertex of $P$ lies on the boundary of $Q$.

Or in a more general version:

Problem 3**.**

Given a convex polytope $P$ in an Euclidean $d$-space, find the maximum volume convex polytopes $Q$ with the property that each $k$-face of $P$ lies on the boundary of $Q$.

We note that a generalization of Theorem 1 to Problems 2 and 3 can be proved easily.

Theorem 5**.**

*Let $P$ be a convex polytope in Euclidean $d$-space, and let $Q$ be a convex polytope such that every $k$-face of $P$ lies on the boundary of $Q$. If $Q$ has maximum volume among such polytopes, then for every facet $F$ of $Q$, $P$ contains the center of gravity of $F$. *

Proof.

Assume for contradiction that the center of gravity $q$ of $F$ does not belong to $P^{\prime}=P\cap F$. Then there is a $(d-2)$-dimensional affine subspace $L$ in $F$ that separates $P^{\prime}$ and $q$, but for which $P^{\prime}\cap L\neq\emptyset$ and $q\notin L$. Let $r_{L,\phi}$ denote the rotation of $\mathbb{R}^{d}$ about $L$ with angle $\phi$, such that for sufficiently small $\phi>0$, $r_{L,\phi}(F)$ intersects $P$. If it exists, let $Q_{\phi}$ denote the convex polytope, obtained by replacing the supporting halfspace $H$ of $Q$ determined by $F$ by $r_{\phi,L}(H)$. Observe that the derivative of $\operatorname{vol}_{d}(Q_{\phi})$ is proportional to the torque of $F$ with respect to $L$. Nevertheless, since $L$ separates $q$ and $F^{\prime}$, this torque is positive, which means that $Q$ has no maximum volume. ∎

It is an interesting question to ask how well the area of a convex polygon $P$ can be approximated by the area of a maximum area circumscribed polygon $Q$. Clearly, $\frac{\operatorname{area}(Q)}{\operatorname{area}(P)}$ can be arbitrarily large. This happens, for example, if the sum of two consecutive angles of $P$ is only slightly larger than $\pi$. On the other hand, $\frac{\operatorname{area}(Q)}{\operatorname{area}(P)}>1$ is satisfied for every convex polygon $P$. The following proposition shows that this ratio can be arbitrarily close to one as well.

Proposition 1**.**

Let $n\geq 6$. Then, for every $\varepsilon>0$ there is a convex $n$-gon $P$ such that for any maximum area polygon $Q$ circumscribed about $P$, we have $\frac{\operatorname{area}(Q)}{\operatorname{area}(P)}<1+\varepsilon$.

Proof.

Let $p_{1},p_{2},\ldots p_{n}$ be the vertices of $P$ in counterclockwise order. For $i=1,2,\ldots,n$, let $T_{i}$ denote the external triangle that belongs to the side $p_{i}p_{i+1}$. We show the existence of a convex $n$-gon $P$ such that $\sum_{i=1}^{n}\operatorname{area}(T_{i})\leq\varepsilon$, and $\operatorname{area}(P)\geq 1$. Since $Q\subset P\cup\left(\bigcup_{i=1}^{n}T_{i}\right)$, this will clearly imply our statement.

Let $p_{1}$, $p_{2}$ and $p_{3}$ be the vertices of a triangle of unit area, in counterclockwise order. We choose the vertex $p_{4}$ in such a way that $p_{4}$ is sufficiently close to $p_{3}$, and $\operatorname{area}(T_{2})<\frac{\varepsilon}{3}$. We choose $p_{n}$ similarly, close to $p_{1}$ and satisfying $\operatorname{area}(T_{1})<\frac{\varepsilon}{3}$. Note that if $p_{4}$ and $p_{n}$ are sufficiently close to $p_{3}$ and $p_{1}$, respectively, then the sum of the areas of the two triangles, one bounded by the lines through $p_{2}p_{3}$, $p_{3}p_{4}$ and $p_{4}p_{n}$, and the other one bounded by the lines through $p_{4}p_{n}$, $p_{n}p_{1}$ and $p_{1}p_{2}$, is less than $\frac{\varepsilon}{3}$. Now if we put the remaining vertices sufficiently close to the segment $p_{4}p_{n}$, then $\sum_{i=1}^{n}\operatorname{area}(T_{i})<\varepsilon$. ∎

Acknowledgements

Markus Ausserhofer acknowledges support through the FWF-project Y782. Zsolt Lángi was supported by the National Research, Development and Innovation Office, NKFIH, K-119670. Géza Tóth was supported by the National Research, Development and Innovation Office, NKFIH, K-111827.