Isoperimetric problem in a sector

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

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=1of 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 D0such 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

A ≤ L

2

2α holds The equality happens iff C is a part of the circle which centers at the vertex O

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

Theorem 2 When π ≤ α ≤ 2π the isoperimetric inequality

A ≤ L

2

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 P0P1, P1P2, , Pn−1Pn

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≤ L

2 n

4n tan(2nα) 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≤ L

2 n

4n tan(4nπ)

holds The equality happens iff O ≡ Pn, PjPj+1 = Ln

n ,Pj\OPj+1 =

α 2n, j =

0, 1, , n − 1

It is the time I go into detail

2 Case 0 < α < π

2.1 Continous case

Let P, Q are two ends of a maximal curve C such that P ∈ Ox, Q ∈ Oz Claim 1: OP = OQ If OP 6= OQ we take ¯P ∈ Ox, ¯Q ∈ Oz such that O ¯P =

O ¯Q, ¯P ¯Q = P Q and changes the curve C to the same shape ¯C which has two ends ¯P , ¯Q It is easy to see that

area(∆OP Q) < area(∆O ¯P ¯Q)

Therefore the curve C encloses the domain which has strictly smaller area than the one enclosed by the curve ¯C It contradict to the assumption of the maximal of the curve C

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

¯

D = {X + re : X ∈ L, −1

2mX ≤ r ≤1

2mX},

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

L The area of D and ¯D are the same, the length of C is bigger than the curve

¯

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 the length Ln=

n

P

j=1

Pj−1Pj tends to the length L of C,

the area An of OP0P1 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≤ L

2 n

4n tan(2nα)

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 show-ing 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

P0P1, P1P2, , Pn−1Pn, 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: P0P1 = P1P2 = · · · = Pn−1Pn If we don’t have this there exists

j ∈ {1, 2, , n} such that Pj−1Pj6= PjPj+1 So Pj is not on the line Lj which orthogonal to Pj−1Pj+1 at its middle point On the line Lj we take the point

¯

Pj such that Pj, ¯Pj lie on the same side with respect to Pj−1Pj+1 and

Pj−1P¯j = Pj+1P¯j =Pj−1Pj+ Pj+1Pj

It is easy to see that

area(∆Pj−1PjPj+1) < area(∆Pj−1P¯jPj+1).

So the curve P0 Pj−1P¯jPj+1 Pnhas 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: P0P2 = P1P3 = · · · = PjPj+2= · · · = Pn−2Pn If we don’t have this there exists j ∈ {1, 2, , n − 2} such that Pj−1Pj+1 6= PjPj+2 From claim

4 we have Pj−1Pj = PjPj+1 = Pj+1Pj+2 so Pj, Pj+1 are not symmetric with respect to the line Kjwhich orthogonal to Pj−1Pj+1at its middle point Taking

X, Y are the symmetric point of Pj, Pj+1(respectively) with respect to Kj and

¯

Pj, ¯Pj+1are the middle points of PjY, Pj+1X (respectively)

It is easy to see that

+) Pj−1Pj+ PjPj+1+ Pj+1Pj+2> Pj−1P¯j+ ¯PjP¯j+1+ ¯Pj+1Pj+2,

+) area(Pj−1PjPj+1Pj+2) < area(Pj−1P¯jP¯j+1Pj+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,

-) P0P1= P1P2= · · · = Pn−1Pn and OP0= OPn

If n is even Pn/2∈ L by the same argument in the continuous case I can reduce

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

P0P1= P1P2= · · · = Pn−1Pn

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= x0cos α, yn = x0sin α

The length of the curve C is

Ln= n

r

P0P2+ · · · + Pn−1P2

n

So we have

L2 n

n =

n−1

X

j=0

[(xj− xj+1)2+ (yj− yj+1)2] (1) The area of the domain D is

An=1

4

n−1

X

j=0

[(xj+ xj+1)(yj+1− yj) + (xj− xj+1)(yj+1+ yj)] (2)

I will use the following discrete Wirtinger’s inequality version

Lemma 1 For all t ∈ (0,2n) we have

2(1 − cos t)

n−1

X

j=1

yj2+ (1 −sin(n − 1)t

sin(nt) )y

2

n ≤

n−1

X

j=1

(yj− yj+1)2+ y21, (3)

2(1 − cos t)

n−1

X

j=1

x2j+ (1 −cos(n − 1)t

cos(nt) )x

2

n ≤

n−1

X

j=1

(xj− xj+1)2− (1 − cos t)x2

0 (4) The equalities holds iff xj = x0cos(jt), yj = r sin(jt), j = 1, 2, , n, r is a positive constant

Proof For t ∈ (0, π

2n) we have sin(jt) > 0, j = 1, 2, , n and cos(jt) > 0, j = 0, 1, 2, , n

So applying Cauchy’s inequality we get

sin(j + 1)t sin(jt) y

2

j + sin(jt) sin(j + 1)ty

2 j+1≥ 2yjyj+1, (5) cos(j + 1)t

cos(jt) x

2

j+ cos(jt) cos(j + 1)tx

2 j+1≥ 2xjxj+1 (6) Summing all j from 1 to n − 1 for (5) and from 0 to n − 1 for (6)

2 cos t

n−1

X

j=1

yj2+sin(n − 1)t sin(nt) y

2

n≥ 2

n−1

X

j=1

yjyj+1,

x20cos t + 2 cos t

n−1

X

j=1

x2j+cos(n − 1)t cos(nt) x

2

n≥ 2

n−1

X

j=0

xjxj+1

Therefore we proved (3), (4)

It is the time to proof Theorem 3 for 0 < α < π/2

Proof Applying Cauchy’s inequality

2 tan( α

2n)(xj+ xj+1)(yj+1− yj) ≥ tan2(α

2n)(xj+ xj+1)

2+ (yj+1− yj)2, (7)

2 tan( α

2n)(xj− xj+1)(yj+1+ yj) ≥ tan2(α

2n)(yj+ yj+1)

2+ (xj+1− xj)2 (8) Noting that x2 = x2

n+ y2

n, y0 = 0 and tan(2nα) > 0 when 0 < α < π/2, so from (2), (7), (8) we have

8Antan( α

2n) ≤ (1−tan

2(α 2n))

n−1

X

j=0

[(xj+1−xj)2+(yj+1−yj)2]+4 tan2(α

2n)

n−1

X

j=0

(x2j+y2j) (9)

Applying Lemma 1 for t =α

n ∈ (0, π

2n) when 0 < α < π/2

4 sin2(α

2n)

n−1

X

j=1

yj2+ (1 −sin(α − t)

sin α )y

2

n≤

n−1

X

j=1

(yj− yj+1)2+ y21,

4 sin2(α

2n)

n−1

X

j=1

x2j+ (1 −cos(α − t)

cos α )x

2

n≤

n−1

X

j=0

(xj− xj+1)2− 2 sin2(α

2n)x

2

0 Because of y0= 0, xn = x0cos α, yn = sin α

4 sin2(α

2n)

n−1

X

j=0

(x2j+ yj2) ≤

n−1

X

j=0

[(xj− xj+1)2+ (yj− yj+1)2] (10)

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

2 cos t(

2k

X

j=0

x2j+

2k

X

j=1

y2j)−

− 2(

2k−1

X

j=0

xjxj+1+ x2kx0cos α +

2k−1

X

j=1

yjyj+1+ y2kx0sin α) ≥ 0, (11)

for n = 2k + 1, 0 < α < π, t = α

2k + 1 In order to prove the inequality (11) we consider the following matrix

H =A B

C 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

A =









2 cos t −1 0 0 0 − cos α

−1 2 cos t −1 0 0 0

0 0 0 · · · − 1 2 cos t −1

− cos α 0 0 0 −1 2 cos t







 ,

B =







0 0 0 0 0 − sin α

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0





 and C =







0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

− sin α 0 0 0 0 0





 ,

D =









2 cos t −1 0 0 0 0

−1 2 cos t −1 0 0 0

0 0 0 · · · − 1 2 cos t −1

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

Ij= det









2 cos t −1 0 0 0 0

−1 2 cos t −1 0 0 0

0 0 0 · · · − 1 2 cos t −1

0 0 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 t , I2=

sin(3t) sin t and

Ij+1= 2Ijcos t − Ij−1

So by induction we are done

All principle minors of H are

I1, I2, , I2k, J1= I2k−1sin2α,

J1I1, J1I2, J1I2k−1, J2= 0

Because of t = α

2k + 1 ∈ (0, π

2k + 1) when o < α < π, all the principle minors

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

3 Case: π ≤ α ≤ 2π

3.1 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 O ¯Q

Because of the curve C lie on one side with respect to K so the new curve P O ¯Q doesn’t cut itself and has the same length with the curve C The new curve

OP ¯Q has two ends which lie on the sector with the center angle π, and encloses 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 O ¯Q

Noting that I can change the curve C a little bit such that I 6≡ O Hence, from the convex of the part which doesn’t contain P, Q of D, the curve P O ¯Q doesn’t cut itself and the area of the domain which is enclosed by P O ¯Q 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, Q0 on the sector Ixz0, 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 new curve P0, Q0 have the same distance to the line OP Let ¯P , ¯Q is on the line