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As we know the origin of isoperimetric problem is the problem confronted by Queen Dido. The problem was to find the shape of the boundary that should be laid down to enclose maximum area. If one assumes a straight coastline, the answer is semicircle. Some years ago, my colleage Ninh Van Thu ask me how is the problem if the coastline is not straight, it likes a sector and two ends of the boundary lie on two sides of the sector. The purpose of this note is giving the answer of Ninh Van Thu’s question
Isoperimetric problem in a sector Dang Anh Tuan, Hanoi University of Science 15/10/2013 Abstract. As we know the origin of isoperimetric problem is the problem confronted by Queen Dido. The problem was to find the shape of the boundary that should be laid down to enclose maximum area. If one assumes a straight coastline, the answer is semicircle. Some years ago, my colleage Ninh Van Thu ask me how is the problem if the coastline is not straight, it likes a sector and two ends of the boundary lie on two sides of the sector. The purpose of this note is giving the answer of Ninh Van Thu’s question. 1 Introduction Ninh Van Thu’s question can be rewritten as following. Given a sector with central angle α. A piecewise smooth curve C has fixed finite length and two ends on two sides of the sector, doesn’ t cut itself. It encloses the domain D. How is the shape of the curve C such that the domain D has maximal area? I will say the above problem is in the continuous case for distinguishing the discrete case which I mean the curve is piecewise linear, consists of finite segments. To give the full answer for continous case I will consider two cases Case 1: 0 < α < π; Case 2: π ≤ α ≤ 2π. 1 I have some notions. The vertex of the given sector is O and the sector is Oxz. In the continuous case: L is the length of the curve C and A is the area of the domain D. In the case 1, it is easy to see that the domain D should be convex. You can imagine like that: if D is not convex I will blow into D such that the length of C doesn’t change and the area of D is strictly bigger. Because of the fixed finite length of C the domain D is always contained in the fixed bounded domain when C changes. So I can use the following Blaschke Selection Theorem: ”Given a sequence {Kn }∞ n=1 of convex sets contained in a bounded set, there is a subsequence {Kmn }∞ n=1 and a convex set K such that Kmn converges to K in the Hausdorff metric.” Because area of D is bounded so we can take supremum of all posible area of D when C changes. There is a sequence of shape of C called Cj enclosing domains Dj whose areas converges to the supremum. Note that I can assume enclosed domains Dj are convex and contained in a bounded domain, so using Blaschke Selection Theorem there is a subsequence of enclosed domains converges to the convex domain D0 such that the area of D0 is the supremum and it is enclosed by a curve whose length is the length of C. Hence there is a shape of C such that D has maximal area. I will call this shape of C is maximal shape and C is a maximal curve. In the next sections I find out the maximal shape should be a part of the circle. In fact I will prove the following isoperimetric inequality. Theorem 1. When 0 < α < π the isoperimetric inequality L2 2α holds. The equality happens iff C is a part of the circle which centers at the vertex O. A≤ In the case 2: the situation is quite different because of non-convexity. In this case I will use some transformation to reduce the case α = π which is the case of Queen Dido. Hence the answer in this case: the maximal shape is semicircle. In this case I have the following isoperimetric inequality. 2 Theorem 2. When π ≤ α ≤ 2π the isoperimetric inequality A≤ L2 2π holds. The equality happens iff C is the semicircle which has a end at the vertex O. When considering the case 1, by approximation I reduce the problem to the discrete case. It is interesting case and I also consider this case for 0 < α ≤ 2π. When 0 < α < π/2 or π ≤ α ≤ 2π the number of segments doesn’t make any trouble. For the case π ≤ α ≤ 2π, I use the method as above. For the case 0 < α < π/2, I use a version of discrete Wirtinger’ inequality. However when π/2 < α < π, I have to divide into case: the number of segments is odd and the number of segments is even. The even case I use symmetry to reduce the case 0 < α < π/2. The odd case I use the following result in matrix theory: ”A symmetric matrix has all nonnegative eigenvalue if all principle minors of it are positive except null determinant.” We denote, in the discrete case, when the number of segments is n : Ln is the length of the curve C consists of n segments P0 P1 , P1 P2 , . . . , Pn−1 Pn where P0 ∈ Ox, Pn ∈ Oz, and An is the area of the domain D. The same as the continuous case, in the discrete case I have the following isoperimetric inequalities Theorem 3. When 0 < α < π the isoperimetric inequality An ≤ L2n α 4n tan( 2n ) holds. The equality happens iff OP0 = OP1 = · · · = OPn , Pj OPj+1 = α/n, j = 0, 1, . . . , n − 1. When π ≤ α ≤ 2π the isoperimetric inequality An ≤ L2n π 4n tan( 4n ) 3 Ln α holds. The equality happens iff O ≡ Pn , Pj Pj+1 = , Pj OPj+1 = ,j = n 2n 0, 1, . . . , n − 1. It is the time I go into detail. 2 2.1 Case 0 < α < π Continous case Let P, Q are two ends of a maximal curve C such that P ∈ Ox, Q ∈ Oz. ¯ ∈ Oz such that OP¯ = Claim 1: OP = OQ. If OP = OQ we take P¯ ∈ Ox, Q ¯ P¯ Q ¯ = P Q and changes the curve C to the same shape C¯ which has two OQ, ¯ It is easy to see that ends P¯ , Q. ¯ area(∆OP Q) < area(∆OP¯ Q). Therefore the curve C encloses the domain which has strictly smaller area ¯ It contradict to the assumption of the than the one enclosed by the curve C. maximal of the curve C. 4 Claim 2: The curve C is symmetric arround the bisector L of Oxy. Indeed, by Steiner symmetrization about the bisector L the domain D become ¯ = {X + re : X ∈ L, − 1 mX ≤ r ≤ 1 mX }, D 2 2 where e is a unit vector which is orthogonal to the bisector L, mX is the length of the intersection between D and the line pass X and is orthogonal to ¯ are the same, the length of C is bigger than the curve L. The area of D and D ¯ ¯ C which encloses D. From the two above claims, I can deduce the problem to the case 0 < α < π/2. That means if I prove Theorem 1 for 0 < α < π/2, for the case 0 < α < π, because of the symmetric of the maximal curve C around the bisector L, considering two subcurve C1 , C2 of C and applying the above result for two sector OxL, OyL, we will be easy to see that the maximal curve C is also a part of the circle which has center at the vertex O and the above isoperimetric inequality. So Theorem 1 is proved for 0 < α < π. For the case 0 < α < π/2, I use approximation method. That is, for each n ∈ N, on a maximal curve C, I take n points P0 ≡ P, P1 , . . . , Pn ≡ Q such that n the length Ln = Pj−1 Pj tends to the length L of C, j=1 5 the area An of OP0 P1 . . . Pn tends to the area A of D as n → ∞. So if Theorem 3 is proved for 0 < α < π/2 that means we have the following isoperimetric inequality An ≤ L2n α 4n tan( 2n ) by taking limit Theorem 1 will be proved for 0 < α < π/2. Therefore, in order to prove Theorem 1 we will prove Theorem 3 for 0 < α < π/2. 2.2 Discrete case Like continuous case, firstly I will use Blaschke Selection Theorem for showing that there is a shape of curve which encloses maximal area domain.The argument here is slightly different because when we take the limit the number of segments may be changed and the linearity may be broken. Fortunately we can pass over it by taking limit of the vertices of C. So by Blaschke Selection Theorem we can get a maximal curve C which consists of n(n ∈ N) segments P0 P1 , P1 P2 , . . . , Pn−1 Pn , P0 ∈ Ox, Pn ∈ Oz. Claim 3: Using similar argument of Claim 1 in continuous case we have OP0 = OPn . If n = 1 Theorem 3 is done. So we assume n ≥ 2. Claim 4: P0 P1 = P1 P2 = · · · = Pn−1 Pn . If we don’t have this there exists j ∈ {1, 2, . . . , n} such that Pj−1 Pj = Pj Pj+1 . So Pj is not on the line Lj which orthogonal to Pj−1 Pj+1 at its middle point. On the line Lj we take the point P¯j such that Pj , P¯j lie on the same side with respect to Pj−1 Pj+1 and Pj−1 Pj + Pj+1 Pj Pj−1 P¯j = Pj+1 P¯j = . 2 It is easy to see that area(∆Pj−1 Pj Pj+1 ) < area(∆Pj−1 P¯j Pj+1 ). 6 So the curve P0 . . . Pj−1 P¯j Pj+1 . . . Pn has the same length Ln with the maximal curve C and enclose the domain whose area is strictly bigger than the area A of D. It is a contradiction to the assumption C is a maximal curve.. If n = 2 by claim 3 and 4, Theorem 3 is done. So we assume n ≥ 3. Claim 5: P0 P2 = P1 P3 = · · · = Pj Pj+2 = · · · = Pn−2 Pn . If we don’t have this there exists j ∈ {1, 2, . . . , n − 2} such that Pj−1 Pj+1 = Pj Pj+2 . From claim 4 we have Pj−1 Pj = Pj Pj+1 = Pj+1 Pj+2 so Pj , Pj+1 are not symmetric with respect to the line Kj which orthogonal to Pj−1 Pj+1 at its middle point. Taking X, Y are the symmetric point of Pj , Pj+1 (respectively) with respect to Kj and P¯j , P¯j+1 are the middle points of Pj Y, Pj+1 X (respectively). It is easy to see that +) Pj−1 Pj + Pj Pj+1 + Pj+1 Pj+2 > Pj−1 P¯j + P¯j P¯j+1 + P¯j+1 Pj+2 , +) area(Pj−1 Pj Pj+1 Pj+2 ) < area(Pj−1 P¯j P¯j+1 Pj+2 ). It is contradiction to the assumption C is a maximal curve. From claim 3, 4, 5 a maximal curve should have following properties: -) it is symmetric arround the bisector L of the given sector, -) P0 P1 = P1 P2 = · · · = Pn−1 Pn and OP0 = OPn . If n is even Pn/2 ∈ L by the same argument in the continuous case I can reduce 7 Theorem 3 for 0 < α < π to Theorem 3 for 0 < α < π/2. 2.2.1 Discrete case: 0 < α < π/2 As above the maximal curve C consists of n segments which has same length P0 P1 = P1 P2 = · · · = Pn−1 Pn and OP0 = OPn . I now put the DesCartesian coordinate such that the origin is the vertex O. Then we have P0 = (x0 , 0), Pj = (xj , yj ), j = 1, 2, . . . , n, xn = x0 cos α, yn = x0 sin α. The length of the curve C is Ln = n So we have L2n = n P0 P12 + · · · + Pn−1 Pn2 . n n−1 [(xj − xj+1 )2 + (yj − yj+1 )2 ]. (1) j=0 The area of the domain D is An = 1 4 n−1 [(xj + xj+1 )(yj+1 − yj ) + (xj − xj+1 )(yj+1 + yj )]. j=0 I will use the following discrete Wirtinger’s inequality version. 8 (2) π Lemma 1. For all t ∈ (0, 2n ) we have n−1 n−1 yj2 + (1 − sin(n − 1)t 2 )yn ≤ (yj − yj+1 )2 + y12 , sin(nt) j=1 x2j + (1 − cos(n − 1)t 2 )xn ≤ (xj − xj+1 )2 − (1 − cos t)x20 . cos(nt) j=1 2(1 − cos t) j=1 n−1 n−1 2(1 − cos t) (3) j=1 (4) The equalities holds iff xj = x0 cos(jt), yj = r sin(jt), j = 1, 2, . . . , n, r is a positive constant. π ) we have Proof. For t ∈ (0, 2n sin(jt) > 0, j = 1, 2, . . . , n and cos(jt) > 0, j = 0, 1, 2, . . . , n. So applying Cauchy’s inequality we get sin(jt) sin(j + 1)t 2 y + y 2 ≥ 2yj yj+1 , sin(jt) j sin(j + 1)t j+1 cos(j + 1)t 2 cos(jt) xj + x2 ≥ 2xj xj+1 . cos(jt) cos(j + 1)t j+1 (5) (6) Summing all j from 1 to n − 1 for (5) and from 0 to n − 1 for (6) n−1 n−1 yj2 + sin(n − 1)t 2 yj yj+1 , y ≥2 sin(nt) n j=1 x2j + cos(n − 1)t 2 xn ≥ 2 xj xj+1 . cos(nt) j=0 2 cos t j=1 n−1 x20 cos t + 2 cos t n−1 j=1 Therefore we proved (3), (4). It is the time to proof Theorem 3 for 0 < α < π/2. Proof. Applying Cauchy’s inequality α α )(xj + xj+1 )(yj+1 − yj ) ≥ tan2 ( )(xj + xj+1 )2 + (yj+1 − yj )2 , (7) 2n 2n α α 2 tan( )(xj − xj+1 )(yj+1 + yj ) ≥ tan2 ( )(yj + yj+1 )2 + (xj+1 − xj )2 . (8) 2n 2n 2 tan( α Noting that x20 = x2n + yn2 , y0 = 0 and tan( 2n ) > 0 when 0 < α < π/2, so from (2), (7), (8) we have n−1 8An tan( n−1 α α α ) ≤ (1−tan2 ( )) [(xj+1 −xj )2 +(yj+1 −yj )2 ]+4 tan2 ( ) (x2 +y 2 ). 2n 2n j=0 2n j=0 j j (9) 9 Applying Lemma 1 for t = α π ∈ (0, ) when 0 < α < π/2 n 2n n−1 4 sin2 ( n−1 α sin(α − t) 2 ) )yn ≤ y 2 + (1 − (yj − yj+1 )2 + y12 , 2n j=1 j sin α j=1 n−1 4 sin2 ( n−1 α cos(α − t) 2 α ) )xn ≤ x2 + (1 − (xj − xj+1 )2 − 2 sin2 ( )x20 . 2n j=1 j cos α 2n j=0 Because of y0 = 0, xn = x0 cos α, yn = sin α n−1 4 sin2 ( α ) (x2 + yj2 ) ≤ 2n j=0 j n−1 [(xj − xj+1 )2 + (yj − yj+1 )2 ]. (10) j=0 From (9), (10) and (1) Theorem 3 is proved for 0 < α < π/2. Therefore, Theorem 1 and Theorem 3 for 0 < α < π/2 and for π/2 ≤ α < π and n is even are proved. 2.2.2 Discrete case: n is odd and 0 < α < π In the proof of Theorem 3 for 0 < α < π/2, we can not do for π/2 ≤ α < π because of Lemma 1 is not true for this case. However we can proof Theorem 3 without using Lemma 1 if we prove the inequality (10) or equivalent to prove the following inequality 2k 2k x2j + 2 cos t( j=0 yj2 )− j=1 2k−1 2k−1 − 2( yj yj+1 + y2k x0 sin α) ≥ 0, xj xj+1 + x2k x0 cos α + for n = 2k + 1, 0 < α < π, t = consider the following matrix (11) j=1 j=0 α . In order to prove the inequality (11) we 2k + 1 H= A C B D where A is matrix of (2k+1)×(2k+1), B is of (2k+1)×(2k), C is of (2k)×(2k+1) and D is of (2k) × (2k) as following 2 cos t −1 0 ...0 0 − cos α −1 2 cos t −1 ...0 0 0 . . . . . . . . . A= , 0 0 0 · · · − 1 2 cos t −1 − cos α 0 0 ...0 −1 2 cos t 10 0 0 0 0 0 0 B= . . . 0 0 0 0 0 0 0 . . . 0 0 − sin α 0 ...0 0 0 .... . . and C = . 0 ...0 0 0 − sin α ...0 0 0 2 cos t −1 0 ...0 0 −1 2 cos t −1 . . . 0 0 . . . . . . . . D= 0 0 0 · · · − 1 2 cos t 0 0 0 ...0 −1 0 0 0 0 . . 0 0 0 0 ...0 0 0 . . . 0 0 0 . . . . . . , . . . 0 0 0 ...0 0 0 0 0 . −1 2 cos t. As we know in the Matrix theory, the inequality (11) holds if all eigenvalues of H are nonnegative or all principle minors of H are positive except its null determinant. So I have to calculate all principle minors of H. Lemma 2. Put 2 cos t −1 Ij = det . 0 0 −1 2 cos t . 0 0 0 −1 . 0 0 ...0 ...0 .... ··· − 1 ...0 0 0 . 2 cos t −1 0 0 . −1 2 cos t the determinant of the square matrix of order j. We have Ij = sin(j + 1)t . sin t Proof. It is easy to see that I1 = sin(2t) sin(3t) , I2 = sin t sin t and Ij+1 = 2Ij cos t − Ij−1 . So by induction we are done. All principle minors of H are I1 , I2 , . . . , I2k , J1 = I2k−1 sin2 α, J1 I1 , J1 I2 , . . . J1 I2k−1 , J2 = 0. π α ∈ (0, ) when o < α < π, all the principle minors Because of t = 2k + 1 2k + 1 of H are positive except null determinant. Hence the inequality (11) is proved. Therefore Theorem 3 is proved for n is odd and 0 < α < π. 11 3 3.1 Case: π ≤ α ≤ 2π Continuous case I will reduce to the case α = π by some transformation. Let L be the bisector of the given sector and K be the line which is orthogonal to L at the vertex O. There are two cases as following figures. Case 1: the curve C and the domain D lie on only one side with respect to K. Case 2: the curve C lies on both side with respect to K. In this case on the side which does not contain two ends P, Q, the part of domain D should be convex. If it is not by blow into the figure we can make the area trictly bigger and the length doesn’t change. In the case 1, I rotate around the vertex the part OQ of the curve C to the ¯ position OQ. 12 ¯ Because of the curve C lie on one side with respect to K so the new curve P OQ doesn’t cut itself and has the same length with the curve C. The new curve ¯ has two ends which lie on the sector with the center angle π, and encloses OP Q the new domain whose area is the same the area of D. So I have reduced the case π < α ≤ 2π to the case α = π. In the case 2, let I be the intersection between the curve C and the line OQ. ¯ I rotate the part OQ of the curve C around I to the position OQ. Noting that I can change the curve C a little bit such that I ≡ O. Hence, ¯ from the convex of the part which doesn’t contain P, Q of D, the curve P OQ ¯ doesn’t cut itself and the area of the domain which is enclosed by P OQ is not smaller than the area of D. So like the case 1, I can reduce the case π < α ≤ 2π to the case α = π. For the case α = π, it is the problem of Queen Dido. So Theorem 2 is proved. Remark. I can use an other transformation to reduce the case π < α ≤ 2π. If α = 2π, I can reduce to the case π < α < 2π as the following figure. The new curve have two ends P, Q on the sector Ixz , which has center angle β ∈ (π, 2π), and the same length with the curve C, encloses the new domain whose area is the same area of D. If π < α < 2π, firstly I rotate around the vertex O such that the new ends of ¯ is on the line new curve P , Q have the same distance to the line OP. Let P¯ , Q 13 ¯ are orthogonal to OP . Secondly, I translate the OP such that P P¯ and Q Q ¯ Q. ¯ new curve P OQ along P P¯ to the position P¯ O ¯ on the line OP , and the same length with The new curve have two ends P¯ , Q the curve C, encloses the new domain whose area is not smaller than the area of D. 3.2 Discrete case By the same argument in the continuous case I can reduce the case π < α ≤ 2π to the case α = π. There is only a note which is the point I rotate around may not be the vertex of C. However it doesn’t make any trouble as you see the following figure. The length of the new curve P0 P1 . . . P¯j−1 P¯j . . . P¯n−1 , P¯n is not bigger than the length of the curve C, and the area of the domain which is enclosed by the new curve is not smaller than the area of D. For the case α = π we can use an other version of discrete Wirtinger’ s inequality like J. Novotna did in [3]. So Theorem 3 is proved for π ≤ α ≤ 2π. 14 Acknowledgments It is a pleasure to thank Ninh Van Thu for his interesting question. I did this note when I was at VIASM. In this time I followed an interesting minicourse of Le Quang Nam where I learnt Blaschke Selection Theorem. Thanks the support of VIASM and the minicourse of Le Quang Nam. I also want to thank Prof. Gerald Folland for his comments. References 1. Fan, K. Taussky, O. Todd, J.: Discrete Analogs of Inequalities of Wirtinger. Monatsh. Math., 59, 1955, pp. 73-90. 2. Milovanovic I. Z ., Milovanovic E. I., Dolicanin D. C ., Mirkovic T. Z., ”New Proofs of Some Discrete Inequalities of Wirtingers type”, Scientific Publications of the State University of Novi Pazar Ser. A: Appl. Math. Inform. and Mech. vol. 5, 1 (2013), 17-21. 3. Novotna J., ”Variations of discrete analogues of Wirtinger’s inequality”, Casopis pro pestovani matematiky, Vol. 105 (1980), No. 3, 278–285. 4. Treibergs A., ”Steiner Symmetrization and Applications”, University of Utah, January 16, 2008. 15 [...]... 14 Acknowledgments It is a pleasure to thank Ninh Van Thu for his interesting question I did this note when I was at VIASM In this time I followed an interesting minicourse of Le Quang Nam where I learnt Blaschke Selection Theorem Thanks the support of VIASM and the minicourse of Le Quang Nam I also want to thank Prof Gerald Folland for his comments References 1 Fan, K Taussky, O Todd, J.: Discrete Analogs... Analogs of Inequalities of Wirtinger Monatsh Math., 59, 1955, pp 73-90 2 Milovanovic I Z , Milovanovic E I., Dolicanin D C , Mirkovic T Z., ”New Proofs of Some Discrete Inequalities of Wirtingers type”, Scientific Publications of the State University of Novi Pazar Ser A: Appl Math Inform and Mech vol 5, 1 (2013), 17-21 3 Novotna J., ”Variations of discrete analogues of Wirtinger’s inequality”, Casopis... encloses OP Q the new domain whose area is the same the area of D So I have reduced the case π < α ≤ 2π to the case α = π In the case 2, let I be the intersection between the curve C and the line OQ ¯ I rotate the part OQ of the curve C around I to the position OQ Noting that I can change the curve C a little bit such that I ≡ O Hence, ¯ from the convex of the part which doesn’t contain P, Q of D, the curve... Q ¯ new curve P OQ along P P¯ to the position P¯ O ¯ on the line OP , and the same length with The new curve have two ends P¯ , Q the curve C, encloses the new domain whose area is not smaller than the area of D 3.2 Discrete case By the same argument in the continuous case I can reduce the case π < α ≤ 2π to the case α = π There is only a note which is the point I rotate around may not be the vertex... determinant So I have to calculate all principle minors of H Lemma 2 Put 2 cos t −1 Ij = det 0 0 −1 2 cos t 0 0 0 −1 0 0 .0 0 ··· − 1 0 0 0 2 cos t −1 0 0 −1 2 cos t the determinant of the square matrix of order j We have Ij = sin(j + 1)t sin t Proof It is easy to see that I1 = sin(2t) sin(3t) , I2 = sin t sin t and Ij+1 = 2Ij cos t − Ij−1 So by induction we are done All... itself and the area of the domain which is enclosed by P OQ is not smaller than the area of D So like the case 1, I can reduce the case π < α ≤ 2π to the case α = π For the case α = π, it is the problem of Queen Dido So Theorem 2 is proved Remark I can use an other transformation to reduce the case π < α ≤ 2π If α = 2π, I can reduce to the case π < α < 2π as the following figure The new curve have two... ends P, Q on the sector Ixz , which has center angle β ∈ (π, 2π), and the same length with the curve C, encloses the new domain whose area is the same area of D If π < α < 2π, firstly I rotate around the vertex O such that the new ends of ¯ is on the line new curve P , Q have the same distance to the line OP Let P¯ , Q 13 ¯ are orthogonal to OP Secondly, I translate the OP such that P P¯ and Q Q ¯ Q ¯... case α = π by some transformation Let L be the bisector of the given sector and K be the line which is orthogonal to L at the vertex O There are two cases as following figures Case 1: the curve C and the domain D lie on only one side with respect to K Case 2: the curve C lies on both side with respect to K In this case on the side which does not contain two ends P, Q, the part of domain D should be convex... of C However it doesn’t make any trouble as you see the following figure The length of the new curve P0 P1 P¯j−1 P¯j P¯n−1 , P¯n is not bigger than the length of the curve C, and the area of the domain which is enclosed by the new curve is not smaller than the area of D For the case α = π we can use an other version of discrete Wirtinger’ s inequality like J Novotna did in [3] So Theorem 3 is proved... blow into the figure we can make the area trictly bigger and the length doesn’t change In the case 1, I rotate around the vertex the part OQ of the curve C to the ¯ position OQ 12 ¯ Because of the curve C lie on one side with respect to K so the new curve P OQ doesn’t cut itself and has the same length with the curve C The new curve ¯ has two ends which lie on the sector with the center angle π, and ... the area of the domain D The same as the continuous case, in the discrete case I have the following isoperimetric inequalities Theorem When < α < π the isoperimetric inequality An ≤ L2n α 4n tan(... the State University of Novi Pazar Ser A: Appl Math Inform and Mech vol 5, (2013), 17-21 Novotna J., ”Variations of discrete analogues of Wirtinger’s inequality”, Casopis pro pestovani matematiky,... area of D when C changes There is a sequence of shape of C called Cj enclosing domains Dj whose areas converges to the supremum Note that I can assume enclosed domains Dj are convex and contained