Extremal convex polygons inscribed in a given convex polygon
EEXTREMAL CONVEX POLYGONS INSCRIBED IN A GIVENCONVEX POLYGON
CSENGE LILI K ¨ODM ¨ON AND ZSOLT L ´ANGI
Abstract.
A convex polygon Q is inscribed in a convex polygon P if everyside of P contains at least one vertex of Q . We present algorithms for findinga minimum area and a minimum perimeter convex polygon inscribed in anygiven convex n -gon in O ( n ) and O ( n ) time, respectively. We also investigateother variants of this problem. Introduction
Motivated by a problem in statistics, in a recent paper [4] Ausserhofer, Dann,T´oth and the second named author examined the algorithmic aspects of findingconvex polygons of maximal area, circumscribed about a given convex polygon.The aim of our paper is to continue this investigation.Our primary goal is to find, among the convex polygons inscribed in a givenconvex n -gon, one with minimum area or perimeter. For these problems we givealgorithmic solutions requiring O ( n ) and O ( n ) steps, respectively. We will see thatthe first problem is relatively easy to solve, but the second one is not. This problemcan be regarded as a variant of the problem of finding the shortest closed billiardtrajectories in a given convex polygon. We note that this problem, proposed also forconvex bodies in general, is an extensively studied area of research closely related,among other things, to dynamical systems and symplectic geometries. For moreinformation on this subject, the reader is referred to the papers [1, 3, 6, 10], the book[15], or the video recording of the highly interesting talk [2] of Artstein-Avidan.Besides the algorithms, following [4], to any minimum area or perimeter convexpolygon inscribed in a convex n -gon, we assign a sequence from { U, N } n describingits combinatorial properties, and completely characterize the sequences that areassigned to some such polygon (for more details, see the first paragraphs of Subsec-tions 2.2 and 3.2). We also collect observations about the properties of inscribed orcircumscribed convex polygons minimizing or maximizing, respectively, some othergeometric quantity.Finally, we remark that (without completeness) the algorithmic aspects of find-ing convex polygons with maximum/minimum area or perimeter in point sets orpolygons under different geometric constraints were studied, e.g. in [7, 9, 12, 13]. Mathematics Subject Classification.
Key words and phrases. convex polygon, perimeter, area, billiard, dual billiard.The second author is supported by the National Research, Development and Innovation Office,NKFI, K-119670, the J´anos Bolyai Research Scholarship of the Hungarian Academy of Sciences,and the BME IE-VIZ TKP2020 and ´UNKP-20-5 New National Excellence Programs by the Min-istry of Innovation and Technology. a r X i v : . [ m a t h . M G ] J a n C.L. K ¨ODM ¨ON AND Z. L´ANGI
We start with the main definition of our paper, which can be regarded as the‘dual’ of Definition 1 of [4]. Here and throughout the paper by area( K ) andperim( K ) we denote the area and the perimeter of the convex region K , respec-tively. Definition 1.
Let C be a convex polygon. If Q is a convex polygon such thatevery side of C contains at least one vertex of Q , we say that Q is inscribed in C .Furthermore, we set(1) a ( C ) = inf { area( Q ) : Q is inscribed in C } , and(2) p ( C ) = inf { perim( Q ) : Q is inscribed in C } . Note that there is an inscribed polygon of arbitrarily small area in C if andonly if C is a convex n -gon with n ≤
4. Thus, to avoid degenerate configurations,throughout this paper C always denotes a convex n -gon with n ≥
5, and vertices p , p , . . . , p n in counterclockwise order. We extend the indices to all integers sothat they are understood modulo n ; i.e. p i = p j if and only if i ≡ j mod n .The structure of the paper is as follows. In Section 2 we find the minimum areapolygons inscribed in C . In Section 3 we consider minimum perimeter polygonsinscribed in C . Finally, in Section 4 we collect our results about circumscribedpolygons which maximize some geometric quantity. In our investigation, for anypoints x, y ∈ R , we denote by xy the closed segment with endpoints x, y , and thelength of xy by | xy | . We regard points as position vectors, and thus, by y − x wemean the vector pointing from x to y . We denote the convex hull of a set X byconv( X ), and for brevity, we call the relative interior points of a segment interiorpoints . 2. Minimum area convex polygons inscribed in C An algorithmic solution.
First, we describe the geometric background forour algorithm.
Theorem 1.
Let Q be a minimum area convex polygon inscribed in C , with vertices q , q , . . . , q k in counterclockwise order. Then the following holds. (i) Q has no two consecutive vertices that are interior points of some sides of C . (ii) If q j is a vertex of Q contained in the interior of p i p i +1 , then the verticesof Q adjacent to q j are p i − and p i +2 , and p i − p i +2 is parallel to p i p i +1 . (iii) There is a minimum area convex polygon Q inscribed in C , with vertices q (cid:48) , q (cid:48) , . . . , q (cid:48) k in counterclockwise order, such that – q j is a vertex of C if and only if q j = q (cid:48) j , and – if q j is an interior point of p i p i +1 , then q (cid:48) j ∈ { p i , p i +1 } .Proof. First, we prove (i). For contradiction, assume that q j and q j +1 are twoconsecutive vertices of Q , and q j and q j +1 are interior points of p i p i +1 and p i +1 p i +2 ,respectively. Let q j − denote the vertex of Q adjacent to q j and different from q j +1 ,and similarly, let q j +2 denote the vertex of Q adjacent to q j +1 and different from q j . If p i p i +1 is not parallel to q j − q j +1 , then one can slide q j on p i p i +1 in a suitable NSCRIBED POLYGONS 3 direction to decrease the area of Q . Thus, it follows from the minimality of the areaof Q that p i p i +1 and q j − q j +1 are parallel. The property that p i +1 p i +2 and q j q j +2 are parallel are obtained by a similar argument. Now, let Q (cid:48) be a convex polygonobtained from Q by replacing q j by any point q (cid:48) j of p i p i +1 . Then, by our previousobservation, Q (cid:48) is a minimum area convex polygon inscribed in C . On the otherhand, q (cid:48) j q j +2 and p i p i +1 are not parallel, which implies that we may slide q j +1 on p i p i +2 in a suitable direction to obtain a convex polygon inscribed in C with areasmaller than area( Q ), which contradicts our assumption.Now we prove (ii). Let q j − and q j +1 denote the vertices of Q adjacent to q j such that q j − ∈ p i − p i and q j +1 ∈ p i +1 p i +2 . Since q j − is not an interior pointof p i − p i , we have q j − ∈ { p i − , p i } . On the other hand, if q j − = p i , then theconvex hull Q (cid:48) of all vertices of Q but q j is a convex polygon inscribed in C witharea( Q (cid:48) ) < area( Q ). Thus, we have q j − = p i − . The equality q j +1 = p i +2 followsby a similar argument. The fact that p i − p i +2 is parallel to p i p i +1 is obtained byrepeating the argument in the previous paragraph. Finally, (iii) is a straightforwardconsequence of (i) and (ii). (cid:3) The following converse of (iii) of Theorem 1 is easy to prove.
Remark 1.
Let Q be a minimum area convex polygon inscribed in C , with vertices q , q , . . . , q k in counterclockwise order, such that every vertex of Q is a vertex of C . Let 1 ≤ s < s < . . . < s m ≤ k such that for every value of t , | s t +1 − s t | ≥ t there is some index i t such that q s t − = p i t − , q s t +1 = p i t +2 and q s t ∈ { p i t , p i t +1 } . Assume that for t = 1 , , . . . , m , p i t p i t +1 is parallel to p i t − p i t +2 , and let q (cid:48) s t be an arbitrary point p i t p i t +1 . For any s / ∈ { s , . . . , s m } , set q (cid:48) s = q s . Then the convex hull Q (cid:48) of the points q (cid:48) , q (cid:48) , . . . , q (cid:48) k is a minimum areaconvex polygon inscribed in C .By Theorem 1 and Remark 1, for any convex polygon C there is a minimumarea polygon Q inscribed in C with the property that every vertex of Q is a vertexof C , and by determining all minimum area polygons with this additional propertyone can determine all minimum area polygons not satisfying this property. Thus,in the following we consider only convex hulls of subsets of the vertex set of C .Let Q be such a minimum area convex polygon. Then every side of Q is either aside or a diagonal of C . These diagonals of C must lie between vertices separatedby a single vertex, as each side of Q must contain at least one vertex of C . Let T i denote the area of the triangle with vertices p i − , p i , p i +1 . Then the problem ofminimizing the area of Q is equivalent to the problem of choosing some elementsof the set { T , . . . , T n } whose sum is maximal under the restriction that no twoelements with consecutive indices are chosen mod n . We denote this maximal valueby A , and present an algorithm that finds the value of A and a subsequence withsum equal to A .First, we compute the values of all T i s. Note that computing the areas of n triangles, using suitable determinants, can be done in O ( n ) steps. Let us dividethe possible subsequences S into two types: If S contains T we say that it is ofType 1, and otherwise it is of Type 2. For any 1 ≤ k ≤ n −
1, we denote by A k the maximum of the sums of the elements of subsequences of T , T , . . . , T k containing no two elements with consecutive indices but containing T , and forany 2 ≤ k ≤ n we denote by A k the maximum of the sums of the elements of C.L. K ¨ODM ¨ON AND Z. L´ANGI subsequences of T , . . . , T k containing no two elements with consecutive indices.Then, clearly, A = max { A n − , A n } . We find the values A k and A k using a recursivealgorithm.Note that A = T , and A = max { T , T } . Let 3 ≤ k ≤ n −
1, and let S be a subsequence of T , . . . , T k , containing T , with maximal sum and with-out consecutive elements. Furthermore, note that if S does not contain T k , then A k = A k − and otherwise A k = A k − + T k . In other words, we have A k =max { A k − , A k − + T k } . We obtain similarly that A = T , A = max { T , T } ,and A k = max { A k − , A k − + T k } for all 3 ≤ k ≤ n . Finally, observe that both A n − and A n , and thus also a ( C ) = area( C ) − max { A n − , A n } , can be computed in O ( n ) steps. Remark 2.
Our algorithm can clearly be carried out in such a way that we keeptrack of all minimum area inscribed polygons whose vertices are the vertices of C .These polygons are not listed by the algorithm (as their number might be evenmore than linear), but represented in the form of a decision tree.2.2. Combinatorial properties.
As in the algorithm in Subsection 2.1, in thissubsection we investigate minimum area convex polygons inscribed in C with theadditional property that all their vertices are vertices of C . To any such minimumarea polygon Q inscribed in C , one can assign a ‘cyclic’ sequence s C, area ( Q ) ∈{ U, N } n , where the k th element s C, area k ( Q ) of s C, area ( Q ) is U (used) if and only if p k is a vertex of Q . Here, by a cyclic sequence we mean a sequence in which theindices of the elements are understood mod n , and for brevity in this subsection weset s ( Q ) = s C, area ( Q ) and s C, area k ( Q ) = s k ( Q ).Clearly, no sequence s ( Q ) contains two consecutive N s. Indeed, if s k ( Q ) = s k +1 ( Q ) = N , then Q is disjoint from [ p k , p k +1 ], which contradicts the conditionthat Q is inscribed in C . Similarly, s ( Q ) contains no three consecutive U s, since if s k − ( Q ) = s k ( Q ) = s k +1 ( Q ) = U , then the area of Q could be reduced further bynot using p k . Our main result in this subsection is the converse of this observation. Theorem 2.
Let s ∈ { N, U } n with n ≥ . Then the following are equivalent. (i) There is some convex n -gon C with a unique minimum area convex polygon Q such that s ( Q ) = s . (ii) The cyclic sequence s contains no two consecutive N s and no three consec-utive U s.Proof. We only need to prove that (ii) implies (i). Let k denote the number of N sin s , and observe that since n ≥
5, (ii) implies that k ≥ k ≥
3. Let Q be a regular k -gon,let the vertices of Q be q , q , . . . , q k in counterclockwise order, and let g i be themidpoint of q i q i +1 for i = 1 , , . . . , k . Set G = conv { g , g , . . . , g k } . Choose somearbitrary small value ε > Q be G . By the conditionsin (ii), for any 1 ≤ i ≤ k , there are either one or two U s between the two N s in s corresponding to q i and q i +1 . If there is one U between them, we glue an isosceles NSCRIBED POLYGONS 5 triangle to Q with q i q i +1 as its base such that the new vertex q (cid:48) i is closer to g i than ε . Similarly, if there are two U s between the two N s corresponding to q i and q i +1 , we glue a symmetric trapezoid to Q , with q i q i +1 as its base such that thetwo new vertices q (cid:48) i , q (cid:48)(cid:48) i are closer to q i q i +1 than ε , and q (cid:48) i q (cid:48)(cid:48) i is parallel to q i q i +1 andits length is less than | q (cid:48) i g i | = | q (cid:48)(cid:48) i g i | . We carry out this operation for all values of i ,and obtain an n -gon, which we denote by C . Similarly, we denote the convex hullof the points q (cid:48) i and q (cid:48)(cid:48) i by Q . Note that if ε is sufficiently small, removing one pointfrom each pair { q (cid:48) i , q (cid:48)(cid:48) i } , the convex hull Q ∗ of the remaining vertices of Q contains G , and the same statement holds if we replace some of the vertices of Q ∗ with thecorresponding midpoints g i . q q' q q q' q'' q q' q Q q' q'' q' Figure 1.
The polygon C constructed in the proof of Theorem 2for k = 5 and s = N U N U N U U N U N U U .We show that Q is the unique minimum area polygon inscribed in C if ε issufficiently small. First, observe that in this case Q is convex and its area is closeto area( G ). In particular, for sufficiently small values of ε the inequality area( Q ) < area(conv( G ∪ { q i } )) is satisfied for all values of i . On the other hand, if Q (cid:48) is anyconvex polygon inscribed in C with vertices chosen from the vertices of C , then q (cid:48) i , q (cid:48)(cid:48) i or g i belongs to Q (cid:48) for all values of i . Indeed, if there is one U between thetwo N s corresponding to q i and q i +1 , then q (cid:48) i is a vertex of Q (cid:48) , or both q i and q i +1 are vertices of Q (cid:48) , implying that g i ∈ Q (cid:48) . In the opposite case the fact that Q (cid:48) is inscribed in C yields that it contains a point of the segment q (cid:48) i q (cid:48)(cid:48) i . Since everyvertex of Q (cid:48) is a vertex of C , from this q (cid:48) i ∈ Q (cid:48) or q (cid:48)(cid:48) i ∈ Q (cid:48) follows. But by ourprevious observation this implies that G ⊆ Q (cid:48) . Thus, if q i is a vertex of Q (cid:48) for somevalue of i , then area( Q ) < area(conv( G ∪ { q i } )) ≤ area( Q (cid:48) ). On the other hand, ifno q i is a vertex of Q (cid:48) , then the facts that every vertex of Q (cid:48) is a vertex of C and Q (cid:48) is inscribed in C implies that Q (cid:48) = Q . By Theorem 1 and Remark 1, since Q is a unique minimum area polygon inscribed in C with the additional assumptionthat every vertex of Q is a vertex of C , it follows that there is no minimum areapolygon inscribed in C having a vertex in the interior of a side of C . Finally, weclearly have s ( Q ) = s , which yields the assertion for k ≥
3. If k = 2, a similarconstruction proves the statement, where the regular polygon Q is replaced by astraight line segment. (cid:3) C.L. K ¨ODM ¨ON AND Z. L´ANGI Minimum perimeter convex polygons inscribed in C An algorithmic solution.
Let F ( C ) denote the family of minimum perime-ter convex polygons inscribed in C . We start with the description of some propertiesof the elements of F ( C ). We first prove Lemma 1, and note that the property de-scribed in it is well known in the theory of billiards, and it can be proved also viaa simple differential geometric argument. Lemma 1.
Let Q be a convex polygon with minimum perimeter inscribed in C . Let q j − , q j , q j +1 be three consecutive vertices of Q . If q j is an interior point of a side p i p i +1 of C , then Q satisfies the optic reflection law at q j , i.e. the angles ∠ p i q j q j − and ∠ q j +1 q j p i +1 are equal.Proof. The locus of the points in the plane with the property that the sum of theirdistances from q j − and q j is a given constant is an ellipse with q j − and q j +1 asits foci. Let E be the ellipse with q j − and q j +1 as its foci, and containing q i on itsboundary. Since q j minimizes the sum of the distances from q j − and q j +1 amongthe points of the line L through p i p i +1 , L is tangent to E at q j . Thus, the equality ∠ p i q j q j − = ∠ q j +1 q j p i +1 follows from the property of ellipses that the tangent line L at q j bisects the exterior angles of the triangle conv { q j − , q j , q j +1 } at q j [5]. (cid:3) Clearly, by Lemma 1, every vertex of a minimum perimeter inscribed polygon iseither a vertex of C or satisfies the reflection law. Definition 2.
Let Q ∈ F ( C ) be a minimum perimeter convex polygon inscribed in C . Then, by s C, perim ( Q ) ∈ { U, N } n we denote the n -element cyclic sequence whose i th element s C, perim i ( Q ) is U if p i is a vertex of Q , and s C, perim i ( Q ) = N otherwise.For brevity, in this subsection we use the notation s ( Q ) = s C, perim ( Q ) and s i ( Q ) = s C, perim i ( Q ).Note that if s ( Q ) contains exactly k U s, then Q has exactly ( n − k ) vertices. It isalso worth noting that convex polygons inscribed in C whose every vertex satisfiesthe optic reflection law are called Fagnano orbits , and that a necessary conditionfor the existence of a Fagnano orbit in an even-sided polygon can be found in [8]as Lemma 2.In the next theorem, we denote by α i the measure of the angle ∠ p i − p i p i +1 . Theorem 3.
Let s ∈ { U, N } n . Then the following holds. (i) If n is odd or s (cid:54) = n (cid:122) (cid:125)(cid:124) (cid:123) N N . . . N , then there is at most one minimum perimeterpolygon Q ∈ F ( C ) with s ( Q ) = s . (ii) If n is even, s = n (cid:122) (cid:125)(cid:124) (cid:123) N N . . . N , then either there is no Q ∈ F ( C ) with s ( Q ) = s ,or there are infinitely many. In the latter case, if Q , Q ∈ F ( C ) satisfy s ( Q ) = s ( Q ) = s , then all corresponding pairs of sides of Q and Q areparallel (cf. Figure 2).Proof. First, we prove (i).
NSCRIBED POLYGONS 7 C Figure 2.
Minimum perimeter polygons inscribed in a regularhexagon, indicated with dashed lines.Consider the case that s contains at least two U s, and let p i and p j be twovertices of Q corresponding to two consecutive U s in s . We label the vertices of Q in such a way that q i = p i , q i +1 ∈ p i +1 p i +2 , . . . , q j − ∈ p j − p j − and q j − = p j .Let q (cid:48) i denote the reflected copy of q i to the line through p i +1 p i +2 . By the opticreflection law, q (cid:48) i , q i +1 and q i +2 are collinear, and | q (cid:48) i q i +2 | = | q i q i +1 | + | q i +1 q i +2 | .Now, if q (cid:48)(cid:48) i denotes the reflected copy of q (cid:48) i to the line through p i +2 p i +3 , then bythe optic reflection law, q (cid:48)(cid:48) i , q i +2 and q i +3 are collinear, and | q (cid:48)(cid:48) i q i +3 | = | q i q i +1 | + | q i +1 q i +2 | + | q i +2 q i +3 | . Continuing this process, if q ∗ i denotes the point obtainedby subsequently reflecting q i to the lines through p i +1 p i +2 , p i +2 p i +3 , . . . , p j − p j − ,respectively, then | q ∗ i q j − | = (cid:80) j − t = i | q t q t +1 | . Note that q i = p i and q j − = p j , andhence, q ∗ i q j − is depends only on s ( Q ) and does not depend on Q . This yields, inparticular, that the length of the boundary of Q from p i to p j is independent of Q . On the other hand, for any i ≤ t ≤ j −
3, if q ∗ t denotes the point obtained bysubsequently reflecting q t to the lines through p t +1 p t +2 , . . . , p j − p j − , the all q ∗ t slie on [ q ∗ i , q j − ]. Thus, q j − is the intersection point of p j − p j − and q ∗ i q j − , q j − is the intersection point of p j − p j − and the reflected copy of q ∗ i q j − to the linethrough p j − p j − , and the remaining points p t can be obtained in a similar way.We note that this process can be carried out in O ( j − i ) steps, including the checkwhether the obtained points q t indeed lie in the interiors of the corresponding sidesof C . If s contains exactly one U , the same argument can be applied in which thepoints p i and p j coincide, and the part of bd( Q ) between p i and p j is equal tobd( Q ).Consider the case that s = N N . . . N , and let q be the vertex of Q on p n p .For l ∈ { n, } , let p ∗ l denote the point obtained by reflecting p l subsequently aboutthe lines through p p , p p , . . . , p n − p n , respectively. Let q ( τ ) = τ p n + (1 − τ ) p ,and q ∗ ( τ ) = τ p ∗ n + (1 − τ ) p ∗ for any τ ∈ (0 , τ there is exactly one closed polygonal curve Γ( τ )starting at q ( τ ), having subsequent vertices q ( τ ) , . . . , q n − ( τ ) on the lines through p p , . . . , p n − p n , respectively, and returning to q ( τ ) such that each vertex q i ( τ )satisfies the optic reflection law for all 1 ≤ i ≤ n −
1. This curve can be obtainedby taking the straight line segment q ( τ ) q ∗ ( τ ), applying reflections and taking in-tersections. Thus, the length of the curve Γ( τ ) is equal to | q ( τ ) q ∗ ( τ ) | . C.L. K ¨ODM ¨ON AND Z. L´ANGI
A necessary condition for some Γ( τ ) to be the boundary of an element Q ( τ ) ∈F ( C ) is that(a) all its vertices are contained in the interiors of the sides of C ;(b) it has minimal length among all curves Γ( τ ), τ ∈ (0 , I ⊆ (0 ,
1) denote the values of τ such that Γ( τ ) satisfies (a). We need to findthe minimum of the length of Γ( τ ) on I . Since for all values of τ , the length ofΓ( τ ) is | q ( τ ) q ∗ ( τ ) | , it is sufficient to do it for | q ( τ ) q ∗ ( τ ) | . Note that | q ( τ ) q ∗ ( τ ) | = aτ + bτ + c for some a ≥ , b, c, ∈ R . Furthermore, as I is clearly open in [0 , | q ( τ ) q ∗ ( τ ) | on I is a local minimum of | q ( τ ) q ∗ ( τ ) | on (0 , τ ) satisfying (a) and (b), or | q ( τ ) q ∗ ( τ ) | isindependent of τ . Assume the latter. Then an elementary consideration shows thatthe vector q ∗ ( τ ) − q ( τ ) is independent of τ , which yields that the correspondingpairs of sides of Q ( τ ) and Q ( τ ) are parallel for all τ , τ ∈ I , where Q ( τ ) denotesconv(Γ( τ )) for all τ ∈ I .Let Q ( τ ) ∈ F ( C ) for some τ ∈ I . To complete the proof we show that there issome neighborhood W of τ in (0 ,
1) such that if n is even then Q ( τ ) ∈ F ( C ) forany τ ∈ W , and if n is odd, then Q ( τ ) / ∈ F ( C ) for any τ ∈ W \ { τ } . Consider some τ ∈ (0 ,
1) sufficiently close to τ . Then, as I is open, we have τ ∈ I . Imagine abilliard ball at q ( τ ) and push it parallel to q ( τ ) q ( τ ). The ball bounces back fromthe side p p at q ( τ ) by the optic reflection law, and runs parallel to q ( τ ) q ( τ ).Here an elementary computation shows that the signed distance of the parallelsegments q ( τ ) q ( τ ) and q ( τ ) q ( τ ) is the opposite of the signed distance between q ( τ ) q ( τ ) and q ( τ ) q ( τ ). Repeating this consideration for all sides of Q ( τ ), weobtain that the billiard trajectory ends at q ( τ ) if and only if the ball bounces backan odd number of times; that is, if n is even. (cid:3) Remark 3.
The proof of (ii) of Theorem 3 yields a little more: if n is even andthere are infinitely many polygons Q ∈ F ( C ) with s ( Q ) = N N . . . N , then thesepolygons are of the form Q ( τ ) for some subinterval I (cid:48) ⊆ (0 , p i p i +1 the values of τ such that afterreflections the corresponding point lies on p i p i +1 is an interval, and the intersectionof intervals is an interval. Remark 4.
Assume that there is some Q ∈ F ( C ) with s ( Q ) = N N . . . N . Thenbd( Q ) coincides with some Γ( τ ) satisfying the properties in (a) and (b). In theother direction, if some Γ( τ ) satisfies the properties in (a) and (b), then, applyingthe reflection argument as in the second part of the proof of Theorem 3 and usingthe elementary fact that the length of any polygonal path connecting two pointsis at least as large as the distance between the points, it follows that Γ( τ ) is theboundary of some Q ∈ F ( C ) with s ( Q ) = N N . . . N . In particular, this impliesthat all elements Q ∈ F ( C ) with s ( Q ) = N N . . . N , and also their perimeter, can befound in O ( n ) steps. Indeed, carrying out the reflections to p n p , we can computethe function τ (cid:55)→ | q ( τ ) q ∗ ( τ ) | and determine its unique local minimum in O ( n ) steps. Remark 5.
Let C be a regular n -gon, and let Q be the convex hull of the midpointsof the edges of C . From Theorem 3 and Remark 4 it follows that Q is a minimumarea convex polygon inscribed in C . Indeed, with the notation of the proof ofTheorem 3 Q coincides with the boundary of Γ(1 / C , we have that Γ( τ ) and Γ(1 − τ ) are congruent, implying that NSCRIBED POLYGONS 9 the function τ (cid:55)→ | q ( τ ) q ∗ ( τ ) | = | q ∗ ( τ ) − q ( τ ) | is symmetric to 1 /
2. This yieldsthat either q ∗ ( τ ) − q ( τ ) is independent of τ , or it moves on a line perpendicular to q ∗ (1 / − q (1 / /
2) is the minimumof | q ( τ ) q ∗ ( τ ) | for all τ ∈ R . By Remark 4, this implies that Q ∈ F ( C ).In the remaining part of Subsection 3.1, we present an algorithm to find anelement of F ( C ) and its perimeter. Our algorithm is based on the one in [4], withthe necessary modifications.For any i, j with i < j ≤ i + n , let Γ ij be a shortest polygonal curve (cid:83) m − t =1 q t q t +1 such that(i) q = p i , q m = p j , and all vertices of Γ ij are boundary points of C ,(ii) all sides p i p i +1 , p i +1 p i +2 , . . . , p j − p j contain at least one vertex of Γ ij , and(iii) the points q , q , . . . , q m are in this councerclockwise order in bd( C ).Let Π ij denote the length of Γ ij . Clearly, if Q ∈ F ( C ), and s ( Q ) (cid:54) = N N . . . N , thenperim( Q ) = Π i,i + n for some value of i . We present a recursive algorithm whichcomputes Π ij for all i < j ≤ i + n .First, clearly, we have Q ij = p i p i +1 for j = i + 1, and for j = i + 2, Q ij = p i p i +2 . This implies that Π i,i +1 = | p i p i +1 | and Π i,i +2 = | p i p i +2 | for all values of i .Consider some 2 < k ≤ n , and assume that we have computed all values Π st with s < t < s + k . Choose some i, j with j = i + k . We distinguish between k types ofthe shortest polygonal curves Γ ij satisfying the properties in the list in the previousparagraph.Type (0): None of the points p i +1 , p i +2 , . . . , p j − is a vertex of Γ ij .Type ( u ): The point p i + u is a vertex of Γ ij for some 1 ≤ u ≤ k − ij has Type (0), then it has no Type ( u ) for any 1 ≤ u ≤ k −
1. On theother hand, in general Γ ij may have Type ( u ) for more than one distinct value of u . Our algorithm determines the value of Π ij depending on the type of Γ ij .If Γ ij has Type (0), then the vertices q , q , . . . , q m − satisfy the optic reflectionlaw by Lemma 1. Then, by Theorem 3, the existence of a polygonal curve satisfyingthese conditions, and in case of existence the vertices of this curve and its lengthcan be found in O ( k ) steps. Assume that Γ ij has Type ( u ) for some 1 ≤ u ≤ k − ij is the union of some shortest polygonal curves Γ i,i + u and Γ i + u,j , and itslength is Π i,i + u + Π i + u,j . To find the shortest polygonal curves having Type ( u )for all possible values of u , and the length of these curves, we need O ( k ) steps.Starting with k = 3, for every fixed value of k we execute the above procedurefor all 1 ≤ i ≤ n . Then we increase the value of k by one and repeat all steps until k = n . Thus, we obtain the values of Π ij for all i, j with i < j ≤ i + n in O ( n )steps. Indeed, we have seen that for any fixed values i and j = i + k , we can findthe value of Π ij in O ( k ) times, and hence, the estimate O ( n ) follows by executingthis procedure for all values of i and k , and using the inequality k ≤ n .Let p ( C ) = min { Π i,i + n : i = 1 , , . . . , n } . We need to handle the case of theconvex polygons Q ∈ F ( C ) with s ( Q ) = N N . . . N . Nevertheless, by Remark 4, allsuch polygons, if they exist, and their perimeter can be found in O ( n ) steps. Inconclusion, p ( C ), and also an inscribed convex polygon with minimum perimeter,can be found in O ( n ) steps. We note that by (ii) of Theorem 3, if n is even and there is some Q ∈ F ( C ) with s ( Q ) = N N . . . N then there is some Q ∈ F ( C ) with s ( Q ) (cid:54) = N N . . . N . This yieldsthat if n is even then p ( C ) = p ( C ); thus, if we want to calculate only the value of p ( C ) we can skip the last step of the algorithm for n even. Remark 6.
As in case of minimum area polygons (cf. Remark 2), our algorithm canbe carried out in such a way that we keep track of all minimum perimeter inscribedpolygons. These polygons are not listed by the algorithm, but represented in theform of a decision tree.3.2.
Combinatorial properties.
In Subsection 3.1 we have seen that for anyminimum perimeter polygon Q ∈ F ( C ), there is a cyclic sequence s ( Q ) ∈ { U, N } n associated to Q such that for all 1 ≤ i ≤ n , the i th element s i ( Q ) of s ( Q ) is U if p i is a vertex of Q , and s i ( Q ) = N otherwise (cf. Definition 2). Our main goal inthis subsection is to determine the cyclic sequences s ∈ { N, U } n with the propertythat for a suitable convex n -gon C there is some Q ∈ F ( C ) such that s ( Q ) = s .Let s ∈ { U, N } n . Similarly like in Subsection 2.2, if s is associated to some convexpolygon Q ∈ F ( C ) with a suitable choice of C , then it contains no three consecutive U s. Indeed, if three consecutive vertices of C are used, then the perimeter of Q could be further reduced by removing the middle vertex from the vertex set of Q .In other words, no ‘realizable’ cyclic sequence s contains three consecutive U s. Ourmain result is the following. Theorem 4.
A cyclic sequence s ∈ { U, N } n is realizable if and only if s does notcontain three consecutive U s.Proof. Since no realizable sequence contains three consecutive U s, we need to provethat if s ∈ { U, N } n does not contain three consecutive U s, then s is realizable.Without loss of generality, we assume that s (cid:54) = N N . . . N .Let the number of N s in such a sequence s ∈ { U, N } n be k . First, we prove theassertion in the case that k ≥
3. Let C be a regular k -gon of unit edge length,with vertices p , p , . . . , p k , where the indices are understood mod k , and let Q be the convex hull of the midpoints of the sides of C . Let the vertices of Q be m , m , . . . , m k such that m i ∈ p i p i +1 . By Remark 5, if k is odd, then Q is theunique smallest perimeter convex polygon inscribed in C . Furthermore, it followsfrom Remark 3 that if k is even, then for any τ ∈ [0 , Q ( τ ), with vertices q ( τ ) , q ( τ ) , . . . , q n ( τ ) in counterclockwise order, suchthat q ( τ ) = τ p + (1 − τ ) p and q i ( τ ) q i +1 ( τ ) is parallel to m i m i +1 for all values of i . Furthermore, the perimeters of these polygons are equal, F ( C ) = { Q ( τ ) : τ ∈ [0 , } , where Q (1 /
2) = Q , and in the degenerate cases τ = 0 and τ = 1, we have q (0) = q (0), and q (1) = q n (1).We note that since the perimeter of a polygon is a continuous function of itsvertices, any convex polygon inscribed in C whose perimeter is ‘close to’ p ( C ) is‘close to’ an element of F ( C ). Moreover, in the family of convex k -gons for anyfixed value of k , p ( C ) is a continuous function of C , implying that if C and C areconvex k -gons and C is ‘close to’ C , then p ( C ) is ‘close to’ p ( C ).Let ζ > n -gon C (cid:48) in the following way. A side S i = p i p i +1 , i = 1 , , . . . , k of C has Type (t) for some t ∈ { , , } , if the i th and the ( i + 1) st N s in s are NSCRIBED POLYGONS 11 separated by t U s. If S i has Type (0), we regard S i as a side of C (cid:48) . If S i hasType (1), we regard m i as a vertex, and the segments S (cid:48) i = p i m i , S (cid:48)(cid:48) i = m i p i +1 assides of C (cid:48) . Assume that S i has Type (2). Then we choose two points p (cid:48) i and p (cid:48)(cid:48) i on S i symmetric to m i such that p i , p (cid:48) i , p (cid:48)(cid:48) i , p i +1 are in this linear order on S i , and | p (cid:48) i p (cid:48)(cid:48) i | = ζ . We regard p (cid:48) i , p (cid:48)(cid:48) i as vertices and p i p (cid:48) i , p (cid:48) i p (cid:48)(cid:48) i , p (cid:48)(cid:48) i p i +1 as sides of C (cid:48) . We calla (possibly degenerate) convex polygon Q (cid:48) a polygon inscribed in C (cid:48) if every side of C (cid:48) contains at least one vertex of Q (cid:48) . By continuity it follows that if ζ is sufficientlysmall, then no minimum perimeter polygon inscribed in C (cid:48) contains a vertex of C .On the other hand, applying the idea of the proof of Lemma 1, we obtain thatany such polygon contains all the p (cid:48) i s and p (cid:48)(cid:48) i s on the Type (2) sides as well as themidpoints of the Type (1) sides of C as vertices. Thus, there is a unique minimumperimeter polygon inscribed in C (cid:48) and the vertex set of this polygon consists of all p (cid:48) i s and p (cid:48)(cid:48) i s on the Type (2) sides, the midpoints of the Type (1) sides of C , andone point in the interior of each Type (0) side of C .Now let u i be the outer unit normal vector of S i for all values of i . Considersome δ >
0. If S i is of Type (1), set ¯ p (cid:48) i = m i + εu i , and if S i is of Type (2), set¯ p (cid:48) i = p (cid:48) i + εu i and ¯ p (cid:48)(cid:48) i = p (cid:48)(cid:48) i + εu i . Let C denote the convex hull of the union of C and all points ¯ p (cid:48) i and ¯ p (cid:48)(cid:48) i . Then, by continuity and Lemma 1, if δ > C is a convex polygon, and any minimum perimeter convex polygon Q inscribed in C has all ¯ p (cid:48) i s and ¯ p (cid:48)(cid:48) i s as vertices, and no vertex of C is a vertex of C . Clearly, the sequence assigned to any such polygon is s , and thus, Theorem 3yields that such a Q is unique. p p' p p'' p' p' p p' p p'' C C p Figure 3.
The polygons C and C constructed in the proof ofTheorem 4. The parts of the boundaries of the inscribed polygons Q and Q in the interiors of the polygons are denoted by dottedlines.We are left with the case that the number k of N s in s is at most 2. By ourconditions, from this we have that s = N U U N U or s = N U U N U U . In these cases,we use the fact that for an obtuse isosceles triangle C with base p p and apex p , the shortest closed polygonal curve containing a point from each side of C isa degenerate ‘double segment’ containing p and the midpoint of p p . By slightlymodifying C as in the previous consideration, we obtain a convex pentagon C anda convex hexagon C such that F ( C i ) consists of a unique element Q i for i = 1 , s ( Q ) = N U U N U and s ( Q ) = N U U N U U (see Figure 3). (cid:3) Additional results about circumscribed polygons
Similarly like in [4], we may consider the problem of maximizing a geometricquantity among convex polygons circumscribed about a given convex polygon. Forthis purpose, let us recall Definition 1 from [4].
Definition 3.
Let C ⊂ R be a convex n -gon. If Q is a convex m -gon that contains C and each vertex of C is on the boundary of Q , then we say that Q is circumscribed about C .As in [4], we intend to exclude the existence of unbounded convex polygonalregions Q , containing C , with the property that each vertex of C lies on a side of Q . Thus, we assume that the sum of any two consecutive angles of C is greaterthan π . In particular, from this it follows that n ≥
5. In this section we collect ourobservations about convex polygons circumscribed about C , and having maximalperimeter or maximal diameter.4.1. Maximum perimeter convex polygons circumscribed about C . Set P ( C ) = sup { perim( Q ) : Q is circumscribed about C } , and note that by our condi-tions, P ( C ) exists, and it is attained by a convex polygon Q . Our main result is asfollows. Theorem 5.
Let Q be a maximum perimeter convex polygon circumscribed about C . Let the vertices of Q in counterclockwise order be q , q , . . . , q m , with the verticesunderstood mod m . For all values of i , let the measure of the angle of Q at q i bedenoted by β i . Then every side q i q i +1 of Q contains at least one and at most twovertices of P , and if p j is the unique vertex of P on q i q i +1 , then (3) | q i p j | cot ( β i ) = | q i +1 p j | cot ( β i +1 ) . Proof.
Note that if q i q i +1 contains no vertex of P , then the remaining sidelines of Q are the sidelines of a convex polygon Q (cid:48) circumscribed about C and satisfyingthe inequality perim( Q ) < perim( Q (cid:48) ). This shows that q i q i +1 contains at least onevertex of C , and a similar argument excludes the case that q i q i +1 contains a uniquevertex of C coinciding with q i or q i +1 . The fact that any such side contains at mosttwo vertices of P follows from the convexity of P . Thus, we may assume that q i q i +1 contains exactly one vertex p j of C , which is in the interior of q i q i +1 .We show that in this case (3) holds. Let L i − , L i and L i +1 denote the linesthrough q i − q i , q i q i +1 and q i +1 q i +2 , respectively. Let d i − and d i +1 denote thedistances of p j from L i − and L j +1 , respectively.Consider the case that L i − and L i +1 intersect, and their intersection point x i isseparated from C by L i . Let the measure of the angle ∠ q i x i q i +1 be denoted by γ i (cf.Figure 4). An elementary computation shows that | q i p j | + | p j q i +1 | = d i − sin β i + d i +1 sin β i +1 .Now, let us change Q by rotating L i around p j . Then the above expression canbe regarded as a function f ( β i ), with β i +1 = π + γ i − β i , and with the values of d i − , d i +1 and γ i fixed. By differentiating f , we obtain that it is a strictly concavefunction of β i , and its unique maximum is attained if 0 = d i − cos β i sin β i − d i +1 cos β i +1 sin β i +1 = | q i p j | cot ( β i ) − | q i +1 p j | cot ( β i +1 ).If L i − and L i +1 do not intersect, or their intersection point is not separatedfrom C by L i , a similar argument can be applied. (cid:3) NSCRIBED POLYGONS 13 p j q i- q i q i+ q i+ L i+ L i L i- i+ ii d i+ d i- x i Figure 4.
Notations for the proof of Theorem 5.In Remark 7 we use the notation in Theorem 5.
Remark 7.
Consider some point q i such that the line L through q i p j supports C .Then it can be shown that L contains at most one point q i +1 satisfying the conditionin (3), and the coordinates of this point can be computed from the coordinates of q i and the vertices of C in O (1) steps. Nevertheless, since the condition determiningthis point seems more complicated than in Sections 2 and 3, to characterize theconvex polygons circumscribed about C and satisfying the conditions in Theorem 5seems to be a more difficult problem than in the cases investigated in this paper.4.2. Maximum diameter convex polygons circumscribed about C . Com-pared to the case of perimeter, it seems much easier to find the maximum diameterof the convex polygons circumscribed about C . Indeed, let x i denote the intersec-tion point of the lines through p i − p i and p i +1 p i +2 , and note that q i exists by ourconditions for C . Then any circumscribed polygon can be obtained as the convexhull of some points q i ∈ conv { p i , p i +1 , x i } , i = 1 , , . . . , n . Thus, the diameter ofany convex polygon circumscribed about C is less than or equal to the diameter D ( X ) of X = conv { x i : i = 1 , , . . . , n } . Here the endpoints of any diameter of X are vertices of X . On the other hand, if X has a diameter xy whose endpoints x, y are not consecutive vertices of X , then there is a convex polygon Q circum-scribed about C whose diameter is D ( X ). Hence, to find the maximum diameterof the convex polygons circumscribed about C , in ‘many’ cases it is sufficient tofind D ( X ).We note that D ( X ) can be computed in O ( n ) steps. To do it, first we computethe points x i and then the vertices of X by Graham’s convex hull algorithm [11]. Ifthe points x i are already ordered according to angles in polar coordinates, as in ourcase, the running time of this algorithm is O ( n ). As the last step, the diametersof X can be computed from the vertices of X by the rotating calipers algorithmof Shamos [14]. Nevertheless, to give a complete algorithm to find the maximumdiameter of the convex polygons circumscribed about C , the remaining case, whenthe diameters of X connect consecutive x i s, must also be handled. Acknowledgements.
The authors express their gratitude to Bal´azs Keszegh for many useful comments.
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C.L. K¨odm¨on, Department of Geometry, Budapest University of Technology, EgryJ´ozsef utca 1., Budapest 1111, Hungary
Email address : [email protected] Z. L´angi, MTA-BME Morphodynamics Research Group and Department of Geometry,Budapest University of Technology, Egry J´ozsef utca 1., Budapest 1111, Hungary
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