An algorithm to find maximum area polygons circumscribed about a convex polygon
AAN ALGORITHM TO FIND MAXIMUM AREA POLYGONS CIRCUMSCRIBED ABOUTA CONVEX POLYGON
MARKUS AUSSERHOFER, SUSANNA DANN, ZSOLT L ´ANGI AND G´EZA T ´OTH
Abstract.
A convex polygon Q is circumscribed about a convex polygon P if every vertex of P lies on at leastone side of Q . We present an algorithm for finding a maximum area convex polygon circumscribed about anygiven convex n -gon in O ( n ) time. As an application, we disprove a conjecture of Farris. Moreover, for thespecial case of regular n -gons we find an explicit solution. Introduction
The algorithmic aspects of finding convex polygons under geometric constraints with some extremal propertyhave been studied for a long time. We list just a few examples. Boyce et al. [7] dealt with the problem of findingmaximum area or perimeter convex k -gons with vertices in a given set of n points in the plane. Eppstein etal. [9] presented an algorithm that finds minimum area convex k -gons with vertices in a given set of n pointsin 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 geometricquantity, were also investigated, see e.g. [13]. Maximum area convex polygons in a given simple polygon wereexamined, e.g. in [5, 14]. Algorithms to find polygons with a minimal number of vertices, nested between twogiven convex polygons, were presented in [6]. The authors of [16] examined among other questions the problemof placing the largest homothetic copy of a convex polygon in another convex polygon. For more information ongeometry-related algorithmic questions, see [1]. Definition 1.
Let P ⊂ R be a convex n -gon. If Q is a convex m -gon that contains P and each vertex of P ison the boundary of Q , then we say that Q is circumscribed about P . Set A ( P ) = sup { area( Q ) : Q circumscribed about P } , if it exists. If area( Q ) = A ( P ) and Q is circumscribed about P , then Q is a maximum area polygon circumscribedabout P .Note that A ( P ) exists if and only if the sum of any two consecutive angles of P is greater than π . Inparticular, this means that P has at least five vertices. Furthermore, observe that if Q is a maximum areapolygon circumscribed about a convex polygon P , then every side of Q contains at least one vertex of P . In thefollowing we assume that these properties hold.Given any convex polygon P , our aim is to find a maximum area convex polygon circumscribed about P . Weinvestigate the properties of these polygons and present an algorithm to find them. Our results can be used tobound an integral of a positive convex function. As an application, we bound an integral of the Lorenz curveand 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 areapolygons circumscribed about a convex n -gon. In Section 3 we present an algorithm with O ( n ) running timethat finds A ( P ) and the maximum area polygons circumscribed about P . Suppose that Q is circumscribed about P . Let S , . . . , S n be sides of P in counterclockwise order. We say that S i is “used” by Q if it is on the boundaryof 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 isused and N otherwise. In Section 4 we investigate the following problem: which sequences can be assigned to a Mathematics Subject Classification.
Key words and phrases. circumscribed polygon, area, Gini index. a r X i v : . [ m a t h . M G ] N ov M. AUSSERHOFER, S. DANN, Z. L´ANGI AND G. T ´OTH 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 ⊂ R denotes a convex n -gon, n ≥
5, and the sum of any two consecutive angles of P is greater than π . The vertices of P are denoted by p , p , . . . , p n , in counterclockwise order. We extend theindices to all integers so that indices are understood mod n ; that is, we let p i = p j if i ≡ j mod n . For any i , wedenote 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 − 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 inthe corresponding external triangle of P . Note that if three consecutive vertices of Q lie on the same line, thenremoving the middle vertex does not change the area of Q . Thus, without loss of generality, in our investigationwe deal only with circumscribed polygons without an angle equal to π . This implies, in particular, that if a sideof P is used (i.e. it is contained in the boundary of the circumscribed polygon Q ), then it is contained in a sideof Q . 2. Geometric properties of maximum area circumscribed polygons
Theorem 1.
For any i, j with i ≤ j ≤ i + n , let Q be a convex polygon circumscribed about P of maximalarea containing S j , S j +1 , . . . , S i + n on its boundary: that is, these edges of P are used by Q . Then for every k = i + 2 , i + 3 , . . . , j − either(a) p k is the midpoint of the side of Q containing it,or(b) S k − or S k lies on bd Q .Proof. Assume that neither S k − nor S k lies on bd Q . This clearly implies that p k belongs to exactly one sideof 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 sideof Q immediately after (resp. before) V in the counterclockwise order, let q + = V ∩ V + and q − = V ∩ 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 α in the clockwise direction, denote the resulting line by L (cid:48) . Let T + be the triangle determined by L + , L ,and L (cid:48) , and let T − be the triangle determined by L − , L , and L (cid:48) . Since | p k q + | > | p k q − | and α is very small, | p k q + | α ≈ area( T + ) > area( T − ) ≈ | p k q − | α . Another possible argument is that since | p k q + | > | p k q − | and α isvery small, the reflection of T + about p k contains T − . Thus modifying Q by replacing L by L (cid:48) would increaseits area. This implies that if Q has the maximum area, then p k is the midpoint of V . (cid:3) Remark 1.
From the proof of Theorem 1 it is clear that it is valid for any maximum area polygon circumscribedabout P . Definition 2.
Let 1 ≤ k ≤ n and let q , q , . . . , q k be points in the plane. We say that the polygonal curve C = q q · · · q k satisfies the midpoint property for the index i , if for j = 1 , , . . . , k , the vertex p i + j of P is themidpoint of q j − q j .If k = n and q = q k , that is if C is a closed polygonal curve and it satisfies the midpoint property for someindex 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 (cid:80) nk =1 ( − k p k (cid:54) = 0 , then there is no closed polygonal curve satisfying the midpointproperty. (2.3) If n is even and (cid:80) nk =1 ( − k p k = 0 , then for every q ∈ R , 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 and p n . In addition, the absolute value of the signed area of C is independent of q . AXIMUM AREA CIRCUMSCRIBED POLYGONS 3
Proof.
Let C = q q · · · q n , q = q n be a closed polygonal curve satisfying the midpoint property for the index 0.For every k , q k is the reflection of q k − about p k . Thus setting q := q , we have q = 2 p − q , q = 2 p − p + q ,etc. In particular, since C is closed, we obtain q = 2 (cid:80) nk =1 ( − n − k p k + ( − n q . If n is odd, it follows that q = (cid:80) nk =1 ( − n − k p k , proving (2.1). If n is even, it follows that (cid:80) nk =1 ( − n − k p k = (cid:80) nk =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 area( C ), is independent of q . For any u, v ∈ R ,we denote by | u, v | the determinant of the 2 × u and v as its columns. Since for every k , | q k − , q k | = | q k − , q k − + q k | = 2 | q k − , p k | , we obtain thatarea( C ) = 12 n (cid:88) k =1 | q k − , q k | = n (cid:88) k =1 | q k − , p k | = n (cid:88) k =1 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k − (cid:88) j =1 ( − k − − j p j + ( − k − q, p k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Thus for some function f = f ( p , p , . . . , p n ), we havearea( C ) = f ( p , p , . . . , p n ) + n (cid:88) k =1 | ( − k − q, p k | = f ( p , p , . . . , p n ) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q, n (cid:88) k =1 ( − k p k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , which is independent of q , since (cid:80) nk =1 ( − k p k = 0. (cid:3) Using the same idea, the following variant of Theorem 2 can be proved.
Theorem 3.
Let ≤ k < n − . Let C denote the family of polygonal curves C = q q · · · q k satisfying themidpoint property for the index i such that q lies on the line L i − through S i − and q j lies on the line L i + j +1 through S i + j +1 . (3.1) If L i − and L i + j +1 are not parallel, then C has exactly one element. (3.2) If L i − and L i + j +1 are parallel and L i + j +1 (cid:54) = 2 (cid:80) jk =1 ( − j − k p i + k + ( − j L i − , then C = ∅ . (3.3) If L i − and L i + j +1 are parallel and L i + j +1 = 2 (cid:80) jk =1 ( − j − k p i + k + ( − j L i − , then for every q ∈ L i − there is exactly one polygonal curve C ∈ C that starts at q . Furthermore, the signed area enclosed by p i − q ∪ C ∪ q j , p i + j +2 ∪ (cid:16)(cid:83) n + i − m = i + j +2 S m (cid:17) is independent of the choice of q . Remark 2. • The unique starting point q in (2.1) can be found in O ( n )-time. Same time is required for the computationof 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 ifeach vertex lies in the corresponding external triangle T i of P . Each of these conditions gives three linearconstraints on the starting point q . The constraints can be obtained in O ( n ) steps, and the intersectionof these 3 n 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 , whichyields that Q contains a side of P . • In (3.1) the unique starting point q , the corresponding solution C = q q . . . q j , its convexity propertiesand 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 ≤ 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 , . . . , S i + n = S i lie on the boundary of Q ij . Let A ij = A ij ( P ) be thearea 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 ≤ i + n . It also finds A ( P ) and themaximum area circumscribed polygons about P . M. AUSSERHOFER, S. DANN, Z. L´ANGI AND G. T ´OTH
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 ≤ n and suppose that we already know the value of A i (cid:48) j (cid:48) ( P ) for every i (cid:48) , j (cid:48) with i (cid:48) < j (cid:48) < i (cid:48) + k .Let j = i + k . Consider a polygon Q circumscribed about P such that the sides S j , S j +1 , . . . , S i + n lie on theboundary of Q . We distinguish between k types of such polygons.Type (0): bd Q does not contain any of the sides S i +1 , S i +2 , . . . , S j − .Type ( α ): bd Q contains the side S i + α for some 1 ≤ α ≤ k − Q can have several types, except for Type (0), which excludes the other types. We find the maximumarea 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 , . . . , p j − hasto be the midpoint of the corresponding side of Q . Whether the sides S i and S j are parallel or no, the existenceof Q and its area can be found in O ( k )-time. This follows from Theorem 3 and Remark 2.Type ( α ): By Theorem 1, an optimal solution Q ij is a union of some - by assumption already known - Q i,i + α and Q i + α,j . It contains the sides S j , S j +1 , . . . , S i + n and S i + α , between S i + n and S i + α it has the samevertices and sides as Q i,i + α , and between S i + α and S j it has the same vertices and sides as Q i + α,j . Its area is A ij = A i,i + α + A i + α,j − area( P ). By construction, the convexity of Q i,i + α and Q i + α,j implies that Q ij is convexas well. Since Q ij can be of any Type ( α ), this step requires O ( k )-time.For each fixed k , starting with k = 3, we execute the above procedure for all 1 ≤ i ≤ n . Then we increase thevalue 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 ≤ i + n .This is done in O ( n )-time. Indeed, let k be fixed, 3 ≤ k ≤ n . Only the case j = i + k is unknown. Q can be ofany type, by above all types require O ( k )-time. Executing this for all i , 1 ≤ i ≤ n , requires O ( kn )-time. Nowwe have A ij for i < j ≤ i + k, ≤ i ≤ n with A is ≤ 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 )-time.Once we have A i ( i + n ) ( P ) for all i , we can calculate the maximum area A ≥ ( P ) of a convex polygon circum-scribed about P containing at least one side of P . A ≥ ( P ) = max { A i ( i + n ) : 1 ≤ i ≤ n } .Denote by A (cid:48) = A (cid:48) ( P ) the maximum area of a convex polygon circumscribed about P containing none of thesides of P . By Remark 2, for even n , if there exists a convex solution containing none of the sides of P , thenthere is a solution containing one side of P with the same area, hence A (cid:48) ≤ A ≥ . However, if we would like tolist all maximum area polygons circumscribed about P , we need to execute this step also in case n is even. Foran 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 Remark2. Finally, A ( P ) = max { A (cid:48) ( P ) , A ≥ ( P ) } , so we get the final answer in O ( n ) time. It is clear that we can keeptrack 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 , ifit lies on 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, foreach value of i , s determines whether the condition (a) or (b) of Theorem 1 is satisfied for p i . In this sectionwe show that most n -element sequences are indeed assigned to some maximum area circumscribed polygon for asuitably chosen P . Our aim is to determine which sequences s ∈ { U, N } n are realizable ; that is, which sequencescan appear for some P (and a maximum area convex polygon Q circumscribed about P ). We do not have acomplete description, but we can tell which sequences can appear as subsequences . Theorem 4. (a) Let s ∈ { U, N } k . It is a subsequence of a realizable sequence s (cid:48) ∈ { U, N } n for some n ≥ k ifand only if s does not contain three consecutive U s. AXIMUM AREA CIRCUMSCRIBED POLYGONS 5
Figure 1.
A ‘long’ subsequence containing only Ns as in (i) (b) Let s ∈ { 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 twosubsequences has length at least . Then s is realizable.Proof. For any x, y, z ∈ R , we denote the triangle 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 thetriangle bounded by these three sidelines strictly increases the area of Q while preserving its convexity. Thisproves 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 asa subsequence.(i) For any i ≥ i = p p . . . p i and two halflines, L and L (cid:48) , startingat p and p i , respectively, such that the triangle T i , bounded by L , L (cid:48) and p , p i contains Γ and so thatthe following property is satisfied. Let ∆ i = q q . . . q i be a convex polygonal curve with q ∈ L , q i ∈ L (cid:48) ,and p j ∈ q j q j +1 for 1 ≤ j ≤ i −
1. Assume that the area of the convex polygon p q q . . . q i p i is maximalamong all convex polygons with the above constraints. Then q q . . . q i satisfies the midpoint property;that is, p j is the midpoint of q j q j +1 for 1 ≤ j ≤ i − i ≥ i = p p . . . p i and a point p so that the polygon pp p . . . p i is convex and so that the following property is satisfied.Let L be the half-line of the line through pp , which starts at p and does not contain p , and let L (cid:48) be the half-line of the line through pp i , which starts at p i and does not contain p . Let ∆ i = q q . . . q i be a convex polygonal curve with q ∈ L , q i ∈ L (cid:48) , and p j ∈ q j q j +1 for 1 ≤ j ≤ i −
1. Assume that thearea of the convex polygon pq q . . . q i is maximal among all convex polygons with the above constraints.Then q q . . . q i satisfies the midpoint property; that is, p j is the midpoint of q j q j +1 for 1 ≤ j ≤ i − s into k consecutivesubsequences n , n , . . . , n k consisting only of N s that are separated by either U or by U U .If k ≥
3, we only need (i). Let ¯ P k = r r . . . r k , where r = r k , be a regular k -gon. Sides r i r i +1 of ¯ P k willbe called old sides . Now we add one or two very small sides, called new sides at each vertex of ¯ P k , accordingto the number of U s separating n m − and n m . When we add one new side at r i , we let it have the same angle M. AUSSERHOFER, S. DANN, Z. L´ANGI AND G. T ´OTH
Figure 2.
A ‘long’ subsequence containing only Ns as in (ii)with the two consecutive old sides. When we add two, we allow them to have almost the same angle with thetwo consecutive old sides. Let ¯ P denote the resulting polygon. Let ¯ 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 transformationsuch that the image of the triangle T i in (i) is h m ( T i ) = ¯ T m , and the image of p , p i is the m th old side of ¯ P .Then, replace this side of ¯ P by h m (Γ i ) for all values of m . We obtain a convex polygon P . If the new sidesare sufficiently small, then for any maximum area polygon circumscribed about P these sides are used. On theother hand, from (i) it follows that no other sides of P are used. If k = 2 and say n ≥
4, then we start ourconstruction with a segment and two nonparallel lines through its endpoints. On one side we apply (ii) with asuitable 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 isoscelestriangle [ p , p , y ] with base p p . Let x be an arbitrary interior point of p p , and q be an arbitrary pointof y x , very close to y . Reflect the points y , q and x about p to obtain the points p , q , y , respectively.Reflect also the points y , q and x about p to obtain p (cid:48) , q and y (cid:48) , respectively (see Figure 3). Then we clearlyhave A ( p , p , y ) = A ( p , p , y ) + A ( p , p (cid:48) , y (cid:48) ) = A ( p , p , q ) + A ( p , p , q ) + A ( p , p (cid:48) , q ). Now slightly rotatethe line of p (cid:48) y (cid:48) about q so that it intersects p y (cid:48) at an interior point y . Let p be the intersection point of thisrotated line and the line of p p (cid:48) . By the idea of the proof of Theorem 1, we have A ( p , y , p ) + A ( p , p , y )
0, ∆ = q q q q does not use any of the sides, so it satisfies the conditions. We can proceedsimilarly 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 , p , p , p (cid:48) , q , q , q , y , y and y as in Figure 3, but relabel y as y (cid:48) (see Figure 5).We obtain y by moving y (cid:48) away from p on the line through p y (cid:48) , and set p as the intersection of thelines through y p and y (cid:48) p (cid:48) . Note that in this case we have A ( p , y , p ) > A ( p , y , p ) + A ( p , y (cid:48) , p ) = (cid:80) j =1 A ( p j − , q j , p j ). Now, choose a point y on the line through p y (cid:48) such that y (cid:48) is an interior point of p y , AXIMUM AREA CIRCUMSCRIBED POLYGONS 7
Figure 3.
An illustration of the proof of (i) for the case i = 3 Figure 4.
An illustration of the proof of (i) for the case i = 4and y is very close to y (cid:48) (see Figure 6). Let q be the reflection of q about p . Consider a point p on the linethrough y p such that the distance of p from y is much smaller that from p . Finally, denote the intersectionof the lines through p p and p q by y , and by L and L (cid:48) those half-lines of the lines through p y and p y ,starting at p and p , that intersect and let p be their intersection. Then the polygon pp . . . p is convex.Observe that if y is sufficiently close to y (cid:48) , and p is sufficiently far from p (i.e. A ( p , y , y (cid:48) ) is small and A ( p , p , q ) is large), then A ( p , y , p ) + A ( p , y , p ) < A ( p , y , p ) + A ( p , y , p ) < (cid:80) j =1 A ( p j − , q j , p j ). Itfollows now directly that conditions in (ii) are satisfied by Γ (cid:48) = p . . . p . To prove (ii) for i > (cid:3) Remark 3.
Let P be a regular n -gon with unit circumradius, where n ≥
5. Then an elementary (but tedious)computation yields the following.(1) If 2 | n , then A ( P ) = n tan πn + n πn cos πn tan πn , and the sequence assigned to a maximum area polygoncircumscribed about P is U N U N . . . U N . M. AUSSERHOFER, S. DANN, Z. L´ANGI AND G. T ´OTH
Figure 5.
An illustration of the proof of (ii)
Figure 6.
Another illustration of the proof of (ii)(2) If 4 | n −
1, then A ( P ) = n tan πn + sin πn cos πn (cid:0) n +34 tan πn − tan π n (cid:1) , and the sequence assigned to a maximumarea polygon circumscribed about P is U n − (cid:122) (cid:125)(cid:124) (cid:123) N N . . . N U N U N . . . U N .(3) If 4 | n + 1, then A ( P ) = n tan πn + sin πn cos πn (cid:0) n +14 tan πn − tan π n (cid:1) , and the sequence assigned to a maximumarea polygon circumscribed about P is U n − (cid:122) (cid:125)(cid:124) (cid:123) N N . . . N U N U N . . . U N .5.
An application to statistics
The above content has a connection to the
Gini index , a measure originally used in economics, statistics, andnowadays 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 otherparameter. For a population of size n , with values (say, wealth) x i in increasing order, F i = i/n , S i = (cid:80) ij =1 x j , AXIMUM AREA CIRCUMSCRIBED POLYGONS 9 and L i = S i /S n . Then the function L ( F ) : F i −→ 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 ≤ α ≤
1, let x α denote the α -quantile of a distribution, that is, exactly α portion of thepopulation has wealth less than x α . Let ¯ x be the mean of the distribution. Then the Lorenz curve is the function L ( p ) := x (cid:82) p x α dα for 0 ≤ p ≤
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 G ( L ) = 2 (cid:90) ( p − L ( p )) dp 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 bythe area under the curve for the neutral scenario.) It is 0 in case of “perfect equality” (everybody has the samewealth) and (almost) 1 in case of “perfect inequality” (one person has all the wealth). pL ( p ) n e u t r a lli n e G i n ii nd e x L o r e n z c u r v e pL ( p ) Figure 7.
Gini index from a known (left) and unknown (right) Lorenz curveWe defined the Lorenz curve as a continuous curve. In practice often only points on the Lorenz curve areknown.
Remark 4.
Data for every individual is often not available and only data for groups is accessible. From thatdata 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 seewhich clients could not pay back their loans and the Lorenz curve is then taken as the percentage of defaults inthe worst i groups. A high Gini coefficient indicates that the bank succeeded in discrimating safe clients fromdangerous clients.For both cases above, the real Gini coefficient is not known and upper and lower bounds are of interest. Bythe convexity of a Lorenz curve the best lower bound is attained by the polygonal curve obtained by connectingthe known points on the Lorenz curve. On the other hand, whereas it is easy to see that any maximal areaconvex curve must be piecewise linear, it does not seem easy to find the best upper bound corresponding toany given point set. A conjecture related to this problem, made by Farris [10], states that the maximal value isattained at a convex polygonal curve with the property that each side of it lies on a sideline of the polygonalcurve connecting the given points, or, using our terminology, no sequence associated to any polygonal curvecontains consecutive N s.Our results imply that Farris’ conjecture does not hold. As a specific counterexample, we may take the partof a regular n -gon P n with 8 | n −
4, centered at (0 , ,
0) and contained in the unit square[0 , . From the computations proving Remark 3 it is easy to see that in this case the optimal circumscribedpolygonal curve does not use any sides of P n . Equivalently, using the idea of the proof of Theorem 4, we mayconstruct 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 Giniindex 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 (cid:48) = P ∩ F . Then there is a( d − L in F that separates P (cid:48) and q , but for which P (cid:48) ∩ L (cid:54) = ∅ and q / ∈ L . Let r L,φ denote the rotation of R d about L with angle φ , such that for sufficiently small φ > r L,φ ( F ) intersects P . Ifit exists, let Q φ denote the convex polytope, obtained by replacing the supporting halfspace H of Q determinedby F by r φ,L ( H ). Observe that the derivative of vol d ( Q φ ) is proportional to the torque of F with respect to L . Nevertheless, since L separates q and F (cid:48) , this torque is positive, which means that Q has no maximumvolume. (cid:3) It is an interesting question to ask how well the area of a convex polygon P can be approximated by thearea of a maximum area circumscribed polygon Q . Clearly, area( Q )area( P ) can be arbitrarily large. This happens, forexample, if the sum of two consecutive angles of P is only slightly larger than π . On the other hand, area( Q )area( P ) > P . The following proposition shows that this ratio can be arbitrarily closeto one as well. Proposition 1.
Let n ≥ . Then, for every ε > there is a convex n -gon P such that for any maximum areapolygon Q circumscribed about P , we have area( Q )area( P ) < ε .Proof. Let p , p , . . . p n be the vertices of P in counterclockwise order. For i = 1 , , . . . , n , let T i denote theexternal triangle that belongs to the side p i p i +1 . We show the existence of a convex n -gon P such that (cid:80) ni =1 area( T i ) ≤ ε , and area( P ) ≥
1. Since Q ⊂ P ∪ ( (cid:83) ni =1 T i ), this will clearly imply our statement.Let p , p and p be the vertices of a triangle of unit area, in counterclockwise order. We choose the vertex p in such a way that p is sufficiently close to p , and area( T ) < ε . We choose p n similarly, close to p andsatisfying area( T ) < ε . Note that if p and p n are sufficiently close to p and p , respectively, then the sum ofthe areas of the two triangles, one bounded by the lines through p p , p p and p p n , and the other one boundedby the lines through p p n , p n p and p p , is less than ε . Now if we put the remaining vertices sufficiently closeto the segment p p n , then (cid:80) ni =1 area( T i ) < ε . (cid:3) Acknowledgements
Markus Ausserhofer acknowledges support through the FWF-project Y782. Zsolt L´angi was supported by theNational Research, Development and Innovation Office, NKFIH, K-119670. G´eza T´oth was supported by theNational Research, Development and Innovation Office, NKFIH, K-111827.
AXIMUM AREA CIRCUMSCRIBED POLYGONS 11
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E-mail address : [email protected] Susanna Dann, Institute of Discrete Mathematics and Geometry, Vienna University of Technology, Wiedner Haupt-strasse 8-10, 1040 Vienna, Austria
E-mail address : [email protected] Zsolt L´angi, Dept. of Geometry, Budapest University of Technology and Economics, Budapest, Egry J´ozsef u. 1.,Hungary, 1111, Research Group of Morphodynamics, Hungarian Academy of Sciences
E-mail address : [email protected] G´eza T´oth, Alfr´ed R´enyi Institute of Mathematics, Hungarian Academy of Sciences, Re´altanoda u. 13-15., 1053Budapest, Hungary
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