Body of constant width with minimal area in a given annulus
aa r X i v : . [ m a t h . M G ] A p r BODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVENANNULUS
A. HENROT, I. LUCARDESI
Abstract.
In this paper we address the following shape optimization problem: find theplanar domain of least area, among the sets with prescribed constant width and inradius. Inthe literature, the problem is ascribed to Bonnesen, who proposed it in [3]. In the presentwork, we give a complete answer to the problem, providing an explicit characterization ofoptimal sets for every choice of width and inradius. These optimal sets are particular Reuleauxpolygons.
Keywords: area minimization; constant width; inradius constraint; Reuleaux polygons2010 MSC: 52A10, 49Q10, 49Q12, 52A38.1.
Introduction
Bodies of constant width (also named after L. Euler orbiforms ) are fascinating geometricobjects and a huge amount of literature has been devoted to them. The fact that many openproblems for these objects remain unsolved, in spite of their simple statement, is probably anelement of their popularity. Among known facts, the famous Blaschke-Lebesgue Theorem assertsthat the Reuleaux triangle minimizes the area among plane bodies of constant width, see [2] forthe proof of W. Blaschke or [12] for a more modern exposition and [13] for the original proofof H. Lebesgue and [3] where this proof is reproduced. Let us mention that many other proofswith very different flavours appeared later, for example [1], [5], [8], [7] and [9].A related problem is the following. For any compact set K , the set of points between theinsphere and the circumsphere is called the minimal shell (or the minimal annulus in dimension2) associated with K . For a body of constant width d , it is known, see [6], that the insphereand the circumsphere are centered at the same point that we will choose as the origin in all thepaper. Moreover, the inradius r and the circumradius R satisfy r + R = d. (1.1)Let us also quote the property: the Reuleaux triangle minimizes the inradius among all bodiesof fixed constant width , see [3] or [6]. Now, given an annulus S with inner radius r and outerradius R satisfying r + R = d with a fixed d >
0, it is natural to try to determine the bodies ofconstant width d having S as their associated minimal annulus and having either maximum orminimum volume. A.E. Mayer in [14] has given upper and lower bounds for the areas of planesets of constant width with prescribed minimal annulus. In particular Mayer’s lower boundyields another proof of the Blaschke-Lebesgue Theorem. The maximization problem has beensolved by T. Bonnesen and the result is explained in the book Bonnesen-Fenchel, see [3], pp.134-135. For the minimization problem, T. Bonnesen gave a conjecture and in a short paper[15], A.E. Mayer gave some sketch of proof which was not complete. Let us quote Chakerian-Groemer whose Chapter on Bodies of constant width in the Encyclopedia of Convexity, see [6],is a well-known reference: ”Mayer in [15] gives a sketch of a proof that the minimum area, for aprescribed annulus, is attained by a certain Reuleaux-type polygon, as conjectured by Bonnesen,however a detailed proof does not appear to have been published.” This is the motivation of ourpaper: we wanted to give a correct, complete and modern proof and describe completely thebody of constant width that minimizes the area among bodies having a given minimal annulus(i.e. bodies having a given inradius).Therefore, in this paper we are concerned with the following problem: determine the optimalshape(s) of A ( r ) := min n | Ω | : Ω ⊂ R , (convex) body of constant width w (Ω) = 1 , ρ (Ω) = r o , (1.2) where ρ (Ω) denotes the inradius. Here, without loss of generality, we have set the width w tobe 1 (clearly, for a generic width t , the minimum and the minimizers have to be rescaled by t and t , respectively). Accordingly, the possible values of the inradius ρ run in the closed interval[1 − / √ , / − / √ ∼ .
422 is the inradius of the Reuleaux triangle, whereas theupper bound is trivial from (1.1). For the extremal values of r , the minimizer is known: on onehand, for r = 1 − / √
3, the optimal shape is the Reuleaux triangle, from Blaschke-Lebesguetheorem; on the other hand, for r = 1 /
2, it is clearly the disk of radius 1 / r , the existence is straightforward andfollows by the direct method of the calculus of variations. Proposition 1.1 (Existence) . Let − / √ ≤ r ≤ / . Then the shape optimization problem A ( r ) has a solution. In this paper we give a complete answer to the problem (1.2), providing an explicit charac-terization of the minimizers for every r . Our construction gives, as a byproduct, uniqueness.In order to state the main result, let us denote by r N +1 , N ∈ N , the inradius of the regularReuleaux (2 N + 1)-gon: r N +1 = 1 −
12 cos (cid:16) π N +1) (cid:17) . The sequence { r N +1 } N is increasing and runs from 1 − / √ / Theorem 1.2 (Characterization and uniqueness) . Let − / √ ≤ r < / .If r = r N +1 for some N ∈ N , then the optimal set of A ( r ) is the regular Reuleaux (2 N +1) -gon.In that case A ( r N +1 ) = (2 N + 1) F ( r N +1 , where F is the function defined in (4.2) .If instead r N − < r < r N +1 for some N ∈ N , N ≥ , setting ℓ := 2 arctan (cid:16)p − r ) − (cid:17) , h := π − N − ℓ, the optimal set of A ( r ) is unique (up to rigid motions) and has the following structure: i) it is a Reuleaux polygon with N + 1 sides, all but one tangent to the incircle; ii) the non tangent side has both endpoints on the outercircle and has length a := 2 arcsin ((1 − r ) sin( h )) , its two opposite sides have one endpoint on the outercircle and meet at a point in theinterior of the annulus; moreover, they both have length b := h + ℓ − a the other N − sides are tangent to the incircle, have both endpoints on the outercircle,and have length ℓ .Moreover, in that case A ( r ) = (2 N − F ( r,
0) + F ( r, h ) (with F defined in (4.2) ). To clarify this result, let us show some picture.
Figure 1.
From left to right: optimal shapes for r = 0 . , . , and . . Notice that in the limit as r → r N +1 , the lengths a , b , and ℓ all converge to2 arctan (cid:0)p − r N +1 ) − (cid:1) , which is the length of the sides of the regular Reuleaux (2 N + 1)-gon. Roughly speaking, in (ii), the interior (to the annulus) point gets closer and closer to the ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 3 outercircle and the non tangent arc gets closer and closer to the incircle. More precisely, weshow the following.
Proposition 1.3 (Continuity) . The functions r argmin A ( r ) and r
7→ A ( r ) are continuousin [1 − / √ , / , the former with respect to the complementary Hausdsorff distance. The continuity of r argmin A ( r ) has to be intended “up to rigid motion”, namely for every ǫ > δ > | r − r | < ǫ ⇒ d H (Ω ; Ω ) < δ, for some representative Ω i ∈ argmin A ( r i ) (unique up to rigid motion), where d H ( · ; · ) denotesthe complementary Hausdorff distance (see, e.g. [10] for the definition).We conclude by pointing out that the scope of Theorem 1.2 is twofold: on one hand, it gives acomplete answer to the Bonnesen’s problem; on the other hand, providing a lower bound of thearea in terms of geometric quantities, it might prove useful in other shape optimization problems.The plan of the paper is the following. The existence of minimizers (proof of Proposition1.1) is given in the next section. As already announced above, in order to characterize theminimizers, we first restrict ourselves to the class of Reuleaux polygons. In this framework,optimal shapes are shown to satisfy an optimality condition, that we call rigidity (see Section3). We use as fundamental tool the so called Blaschke deformations (see Section 2). In the lastsection we characterize the optimal rigid shapes (Theorem 4.1) and we show that actually theyare the minimizers of the original problem A (proof of Theorem 1.2). The very end of the paperis devoted to the continuity statement (proof of Proposition 1.3).2. Preliminaries and Blaschke deformations
This section is devoted to some preliminary tools. In the first part, we give the precisedefinition of width and inradius of a set, and we write the proof of Proposition 1.1. In thesecond part, we gather some facts on Reuleaux polygons: more precisely, we recall the notationand the family of deformations introduced by Blaschke in [2], see also [12] for more details, andwe write the first order shape derivative with respect to these particular deformations.
Definition 2.1.
Given a bounded open set Ω ⊂ R and a direction ν ∈ S , we define the width w ν (Ω) of Ω in direction ν as the minimal distance of two parallel lines orthogonal to ν enclosingΩ. We say that Ω has constant width if w ν (Ω) is constant for every choice of ν . In this case, thewidth is simply denoted by w (Ω). The inradius of Ω, denoted by ρ (Ω), is the largest r for whichan open disk of radius r is contained into Ω. We also recall the classical Barbier Theorem, see[6]: the perimeter of any plane body of constant width d is given by P (Ω) = πd . Proof of Proposition 1.1.
By definition the admissible shapes are (strictly) convex and (up totranslations) their boundary lie in the closed circular annulus A := B − r (0) \ B r (0). If Ω n , n ∈ N , is a minimizing sequence for A ( r ), we can extract a subsequence (not relabeled) which,by Blaschke selection theorem, converges for the Hausdorff distance to some convex set Ω ∗ ,whose boundary is in the annulus A . In particular, ρ (Ω ∗ ) ≥ r . Arguing by contradiction, weinfer that the inradius of Ω ∗ has to be r ; similarly, w ν (Ω ∗ ) = 1 for every direction ν ∈ S (itcan also be proved using the support function). Therefore Ω ∗ is an admissible shape. Finally,since the area is continuous with respect to the Hausdorff convergence for convex domains, Ω ∗ is a minimizer for A ( r ), concluding the proof. (cid:3) Reuleaux polygons and Blaschke deformations.
Reuleaux polygons form a particularsubclass of constant width sets (here fixed equal to 1), whose boundary is made of an odd numberof arcs of circle of radius 1. The arcs are centered at boundary points, intersection of pairs ofarcs. We call such centers vertexes and we label them as P k , k = 1 , . . . , N + 1, for a suitable N ∈ N . The arc opposite to P k is denoted by Γ k and is parametrized byΓ k = { P k + e it : t ∈ [ α k , β k ] } , for some pair of angles α k , β k . Here, with a slight abuse of notation, e it stands for (cos t, sin t ) ∈ R . The subsequent and previous points of P k are P k +1 = P k + e iα k and P k − = P k + e iβ k , A. HENROT, I. LUCARDESI respectively. Accordingly, the angles satisfy β k +1 = α k + π mod 2 π. The concatenation of the parametrizations of the arcs provides a parametrization of theboundary of the Reuleaux polygon in counter clock wise sense: the order is Γ N +1 , Γ N − , . . . ,Γ , Γ N , Γ N − , . . . , Γ , namely first the arcs with odd label followed by the arcs with evenlabel. Notice that the length of the arc Γ k is β k − α k and since the perimeter of the Reuleauxpolygon is π by Berbier Theorem, we have P k β k − α k = π . Remark 2.2.
To clarify the notation above, let us see the case of a Reuleaux pentagon. Γ Γ Γ Γ Γ α β P P P P P Figure 2.
A Reuleaux pentagon (here, for simplicity, regular).
In Figure 2, we have chosen, without loss of generality, α = 0, namely the vertex P alignedhorizontally with P . Accordingly, the angles are ordered as follows0 = α < β < α < β < α < β < α < β < α < β < π and β = α + π , α = β + π , β = α + π , α = β + π , β = α + π . We now introduce a family of deformations in the class of Reuleaux polygons of width 1, whichallow to connect any pair of elements in a continuous way (with respect to the complementaryHausdorff distance), staying in the class. This definition has been introduced by W. Blaschke in[2] and analysed by Kupitz-Martini in [12].
Definition 2.3.
Let Ω be a Reuleaux polygon with 2 N + 1 arcs. Let k be one of the indexes in { , . . . , N + 1 } . A Blaschke deformation acts moving the point P k on the arc Γ k − increasingor decreasing the arc length. Consequently, the point P k +1 moves and the arcs Γ k , Γ k +1 , andΓ k +2 are deformed, as in Fig. 3. We say that a Blaschke deformation is small if the arc lengthof Γ k − has changed of ǫ ∈ R , small in modulus.Let us consider a small Blaschke deformation acting on P k as in Definition 2.3, for somesmall ǫ ∈ R . Let us denote by Γ ǫi , P ǫi , α ǫi , and β ǫi the deformed arcs, vertexes, and angles. Bydefinition, α ǫk − = α k − + ǫ, β ǫk = β k + ǫ. (2.1)The dependence on ǫ of the other angles is less evident. However, it can be derived by imposingthat the transformed configuration is a Reuleaux polygon. Let us determine the first orderexpansion in ǫ . In the following, for brevity we use the symbol ∼ to denote an error of order o ( | ǫ | ). The angles α ǫk , β ǫk +1 , and β ǫk +2 are of the form ( α ǫk ∼ α k + ǫτ, β ǫk +1 ∼ β k +1 + ǫτ,α ǫk +1 ∼ α k +1 + ǫσ, β ǫk +2 ∼ β k +2 + ǫσ, (2.2)for some σ , τ ∈ R . The coefficients σ and τ are uniquely determined by the relation P ǫk +1 = P ǫk + e iα ǫk = P k +2 + e iβ ǫk +2 , which, using the expansions (2.1) and (2.2), easily leads to e iα k − + τ e iα k = σe iβ k +2 ⇐⇒ e iα k − − τ e iβ k +1 = − σe iα k +1 ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 5 P εk P k − P k − P εk +1 P k +2 Γ εk − Γ εk +1 Γ εk Γ εk +2 Figure 3.
A Blaschke deformation of a Reuleaux heptagon which moves P k on Γ k − changing α k − into α ǫk − := α k − + ǫ , with ǫ > small. ⇐⇒ e i ( α k − − α k +1 ) − τ e i ( β k +1 − α k +1 ) = − σ ⇐⇒ ( σ = sin( β k − α k ) / sin( β k +1 − α k +1 ) τ = sin( α k − − α k +1 ) / sin( β k +1 − α k +1 ) . (2.3)2.2. Shape derivatives with respect to Blaschke deformations.
In this paragraph wecompute the first order shape derivative of the area at a Reuleaux polygon, with respect to asmall Blaschke deformation. We recall that, given a one parameter family of small deformationsΩ ǫ of Ω, the first order shape derivative of the area at Ω is nothing but the derivative withrespect to ǫ of the map ǫ
7→ | Ω ǫ | evaluated at ǫ = 0, namely the limitlim ǫ → | Ω ǫ | − | Ω | ǫ . In the following we adopt the symbol ∼ to denote an error of order o ( ǫ ), which does not playany role in the computation of the first order shape derivative. Proposition 2.4.
Let Ω be a Reuleaux polygon with angles α i and β i , i = 1 , . . . , N + 1 . Thefirst order shape derivative of the area at Ω with respect to a small Blaschke deformation actingon the point P k is d A B := 1 − cos( β k − α k ) − sin( β k − α k )sin( β k +1 − α k +1 ) (cid:0) − cos( β k +1 − α k +1 ) (cid:1) . that can also be written introducing the lengths j k = β k − α k and j k +1 = β k +1 − α k +1 of the arcs Γ k and Γ k +1 : d A B = 2 sin( j k / j k +1 /
2) sin (cid:18) j k − j k +1 (cid:19) . In particular, the area decreases under a Blaschke deformation if • P k moves on Γ k − in the direct sense ( ǫ > ) and j k < j k +1 , • P k moves on Γ k − in the indirect sense ( ǫ < ) and j k > j k +1 .Moreover, the case where j k = j k +1 corresponds to a local maximum of the area and the areadecreases when P k moves on Γ k − in both senses.Proof. It is well known (see, e.g. [11]) that the first order shape derivative of the area at aLipschitz domain Ω is a boundary integral which only depends on the normal component of thedeformation. More precisely, it reads Z ∂ Ω V · n d H , where V is the vector field such that Ω ǫ = { x + ǫV ( x ) : x ∈ Ω } , or equivalently, ∂ Ω ǫ = { x + ǫV ( x ) : x ∈ ∂ Ω } . For the Blaschke deformation under study, we clearly have V · n k and Γ k +1 (see Definition 2.3). Using the parametrization [ α j , β j ] ∋ t P j + e it A. HENROT, I. LUCARDESI of Γ j , j = k, k + 1, and noticing that the outer normal vector is e it , we immediately have thefollowing simplification: Z ∂ Ω V · n d H = Z Γ k V · n d H + Z Γ k +1 V · n d H = Z β k α k V ( t ) · e it d t + Z β k +1 α k +1 V ( t ) · e it d t, (2.4)where, for brevity, we have denoted by V ( t ) the vector V ( P j + e it ) on the arc Γ j , j = k, k + 1.Recalling the expansions (2.1) and (2.2) of the angles α ǫk and β ǫk , we infer thatΓ ǫk = { P ǫk + e it : t ∈ [ α ǫk , β ǫk ] }∼ { P k + e it + ǫ ( ie iα k − + iC k ( t ) e it ) : t ∈ [ α k , β k ] } , with C k ( t ) := τ + (1 − τ )( t − α k ) / ( β k − α k ) , and τ defined in (2.3).Therefore, V acts on the arc Γ k as V ( t ) = ie iα k − + iC k ( t ) e it . In particular, V ( t ) · e it = sin( t − α k − ) on Γ k . (2.5)Similarly, using the expansions in (2.2) of α ǫk +1 and β ǫk +1 , and recalling the definition of σ in(2.3), we infer that the arc Γ k +1 is transformed intoΓ ǫk +1 = { P ǫk +1 + e it : t ∈ [ α ǫk +1 , β ǫk +1 ] }∼ { P k +1 + e it + ǫ ( iσe iβ k +2 + iC k +1 ( t ) e it ) : t ∈ [ α k +1 , β k +1 ] } , with C k +1 ( t ) = σ + ( τ − σ )( t − α k +1 ) / ( β k +1 − α k +1 ) . Thus, recalling that β k +2 = α k +1 + π modulo 2 π , V ( t ) · e it = − σ sin( t − α k +1 ) on Γ k +1 . (2.6)Inserting (2.5) and (2.6) into (2.4), developing the integral, we get the first formula. Thesecond follows using elementary trigonometry. The conclusion, in the case of equality of thelengths, comes from the fact that the derivative becomes negative in the direct sense when weperform the Blaschke deformation (since j k +1 increases and j k decreases) and vice-versa. (cid:3) Remark 2.5.
For any non regular Reuleaux polygon, we observe that we can always choosea Blaschke deformation such that the first derivative of the area is negative, making the areadecrease. This is precisely the idea used by W. Blaschke in his proof of the Blaschke-LebesgueTheorem. We can also make the area increase (for a non regular Reuleaux polygon), whichimplies the Firey-Sallee Theorem asserting that the regular Reuleaux polygons maximize thearea among Reuleaux polygons with a fixed number of sides, see [6], [12].3.
Rigid shapes
We have seen that a Blaschke deformation allows to make the area decrease. Therefore, forour minimization problem we can concentrate on sets for which no such Blaschke deformationis permitted (because any Blaschke deformation would violate the annulus constraint). This isthe sense of the next definition.
Definition 3.1.
Let r be fixed. We say that a Reuleaux polygon is rigid if no Blaschke defor-mation that decreases the area can be performed keeping the inradius constraint satisfied. Forbrevity, since we are searching for minimizers in the class of Reuleaux polygons with width 1 andinradius r , we will refer to these particular objects simply as rigid shapes or rigid configurations .A Blaschke deformation is impossible in our class of sets if it moves an arc inside the incircle(or outside the outercircle) violating the constraint of minimal annulus. This is why we introducethe following definitions that describe the only possible arcs such that no deformation is possible. Definition 3.2.
Let be given a Reuleaux polygon of width 1 and inradius r . We say that onearc of its boundary is extremal if it is tangent to the incircle and both endpoints are on theoutercircle. ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 7
Definition 3.3.
Let be given a Reuleaux polygon of width 1 and inradius r . We say that threearcs Γ k − , Γ k , and Γ k +1 of the boundary form a cluster if:the arcs Γ k − and Γ k +1 are tangent to the incircle, their common point P k lies in the interiorof the annulus B − r (0) \ B r (0), and the other endpoints P k ± are on the outercircle.Furthermore, we define the characteristic parameter h as half of the angle P k +1 b OP k − . Thesedefinitions are summarized in Fig. 4. OP k P k − P k +2 P k +1 P k − Γ k Γ k − Γ k +1 h Figure 4.
A triple of arcs (Γ k − , Γ k , Γ k +1 ) forming a cluster and the charac-teristic parameter h . Remark 3.4.
In the definition of cluster, the arc Γ k can be arbitrarily close to the empty setor to an extremal arc. In the first limit case, we have that Γ k − and Γ k +1 form a unique arctangent to the incircle, namely an extremal arc. In the second limit case, Γ k − and Γ k +1 area pair of extremal arcs. All in all, extremal arcs (counted individually or in suitable groups ofthree) can be seen as particular cases of clusters. At last, let us remark that any cluster has anaxis of symmetry.The fundamental proposition in our approach is the following. It shows that we can restrictthe study of optimal shapes to Reuleaux polygons having only extremal arcs and clusters. Proposition 3.5.
The boundary of a rigid shape is made of a finite number (possibly zero) ofclusters and of extremal arcs.Proof.
Let us start with some elementary observations. • No vertex can lie on the incircle. • When a vertex is on the outercircle, its corresponding arc is tangent to the incircle andthis arc goes over the tangent point on both sides. • Conversely, when a vertex is in the interior of the annulus, its corresponding arc is nottangent to the incircle.Assume that the set Ω is rigid. First of all, let us prove that if a set has two consecutivevertexes, say P k and P k +1 lying in the interior of the annulus, it cannot be rigid. Indeed, insuch a case the two arcs Γ k and Γ k +1 are not tangent and therefore the Blaschke deformationdescribed in Definition 2.3 is admissible in both senses ( ǫ > ǫ <
0) without violating theannulus constraint. Now, following Proposition 2.4 we see that such a deformation will decreasethe area by choosing ǫ > j k ≤ j k +1 or ǫ < j k ≥ j k +1 .Now let us consider a point P k lying in the interior of the annulus with its two opposite points P k − and P k +1 on the outercircle. We want to prove that these three points belong to a cluster,namely that P k − and P k +2 are on the outercircle. Let us assume, for a contradiction, that P k +2 is in the interior of the annulus (it will obviously be the same proof with P k − ). According tothe beginning of the proof, necessarily P k +3 has to be on the outercircle.In that case, two particular admissible Blaschke deformations can be considered: • Move P k +1 on Γ k in the direct sense (in the direction of P k − ). • Move P k +1 on Γ k +2 in the indirect sense (in the direction of P k +3 ). A. HENROT, I. LUCARDESI
According to Proposition 2.4, the area will decrease for the first deformation as soon as j k +1 ≤ j k +2 , while it will decrease for the second deformation as soon as j k +1 ≤ j k . Therefore, weobtain the conclusion (this configuration is not rigid) if we can prove j k +1 ≤ max( j k , j k +2 ). Thisclaim is proved in the next Lemma. (cid:3) Lemma 3.6.
Assume that P k and P k +2 lie in the interior of the annulus, and that P k − , P k +1 , P k +3 lie on the outercircle. Then the lengths j k , j k +1 , j k +2 of the arcs Γ k , Γ k +1 , Γ k +2 , satisfy j k +1 ≤ max( j k , j k +2 ) .Proof. Let us introduce the two characteristic parameters h k and h k +2 as half of the angles P k +1 b OP k − and P k +3 b OP k +1 , see Fig. 5. Elementary trigonometry provides the following rela- O h k +2 h k P k P k − P k +1 P k +2 P k +3 j k h k O P k P k − P k +1 Figure 5.
The parameters h k and h k +2 . tions with the corresponding lengths j k and j k +2 :(1 − r ) sin h k = sin( j k / , (1 − r ) sin h k +2 = sin( j k +2 / . Now let us write the angle (or length) j k +1 as j k +1 = P k +2 [ P k +1 P k = P k +2 [ P k +1 O + O [ P k +1 P k . In the triangles P k +1 OP k +2 and P k +1 OP k we get the relations P k +2 [ P k +1 O = h k +2 − j k +2 , O [ P k +1 P k = h k − j k . Therefore j k +1 = h k + h k +2 − j k + j k +2 . Now the lengths j k , j k +2 are less than the length of an extremal arc given by ℓ =2 arctan (cid:16)p − r ) − (cid:17) (see Proposition 3.8). We get the thesis if we can prove that fortwo positive numbers x, y ∈ [0 , ℓ ] and for r ∈ [1 − / √ , / (cid:18) sin( x/ − r (cid:19) + arcsin (cid:18) sin( y/ − r (cid:19) − x + y ≤ max( x, y ) . (3.1)Without loss of generality, by symmetry, we can assume y ≥ x , so that the right-hand side in(3.1) is y . Let us introduce the function G ( x, y ) := arcsin (cid:18) sin( x/ − r (cid:19) + arcsin (cid:18) sin( y/ − r (cid:19) − x + 3 y . We have ∂G∂y = 12(1 − r ) cos( y/ q − sin ( y/ − r ) − . Since the function c c q − − c (1 − r ) ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 9 is decreasing (its derivative has the sign of 1 − / (1 − r ) ), the maximum value of the derivative ∂G∂y is obtained for y = ℓ . This implies ∂G∂y ≤ − r ) cos( ℓ/ q − sin ( ℓ/ − r ) −
32 = 12 − − r ) − ≤ , where we have used the expression cos( ℓ/
2) = 1 / (2(1 − r )) and the bound (1 − r ) ≤ / y G ( x, y ) is decreasing and its maximum on the triangle 0 ≤ x ≤ y ≤ ℓ is on theline x = y . Exactly in the same way, it is immediate to check that x G ( x, x ) is decreasing,thus G ( x, y ) ≤ G (0 ,
0) = 0, proving the lemma. (cid:3)
Example 3.7.
Regular Reuleaux polygons are clearly rigid shapes for their inradius. For ageneric r / ∈ { r N +1 } N , many different rigid configurations can be constructed as we will seebelow. For example, there is one (up to rotations) rigid configuration with one cluster, sincein that case the parameter h is fixed (see (3.8) below). When r is large enough, we can finda continuous family of rigid shapes with two clusters. They are characterized by an arbitrarypair of parameters h , h such that their sum h + h is fixed. And similarly for rigid shapeswith more clusters. Actually, as shown by Proposition 3.8 below, the lengths of arcs in a clusterare completely characterized by the parameter h , moreover, the constraint that the sum of alllengths is π fixes the sum of these parameters. Figure 6.
Two rigid configurations for r = 0 . . On the left, the one with asingle cluster, on the right, one with two clusters. In the next proposition we show that the length of an extremal arc is uniquely determined by r , whereas that of a cluster can be expressed as a function of h (which is instead not uniquelydetermined by r , see also Example 3.7). Proposition 3.8.
Let r be fixed. The length of an extremal arc is ℓ ( r ) := 2 arctan (cid:16)p − r ) − (cid:17) . (3.2) Let (Γ k − , Γ k , Γ k +1 ) form a cluster of parameter h . Then the length of the arc Γ k is a ( r, h ) := 2 arcsin((1 − r ) sin( h )) , (3.3) and the length of the opposite arcs Γ k ± is b ( r, h ) := h + ℓ ( r ) − a ( r, h )2 . (3.4) Moreover, b ( r, h ) ≥ a ( r, h ) .Proof. Throughout the proof we omit the dependence on r and h , which are fixed.Let ℓ denote the length of an extremal arc Γ with opposite point P and endpoints Q and R .The triangle P OQ is isosceles, with base of length 1, legs of length 1 − r , and base angle ℓ/ ℓ/
2) = 1 / (2(1 − r )), which gives (3.2).Let us now consider a cluster of parameter h . Without loss of generality, the involved vertexesare P , . . . , P , oriented in such a way that the parameter h is the angle between the verticalline through O and the segment OP , see Fig. 8-left. According to this notation, we have to OP Q − r − rℓ/ Figure 7.
Computation of ℓ . determine the length a of Γ , and the length b of Γ and Γ . Let us consider the triangle P OP ,see Fig. 8-right: the side P P has length 1 and its opposite angle is π − h ; similarly, the side OP has length 1 − r and its opposite angle is a/
2; therefore a is determined by the relationsin( a/
2) = (1 − r ) sin h , which implies (3.3). Let us now compute b . It is the sum of two angles: O hP P P P P a bb O h a − r P P Figure 8.
Left: cluster configuration under study. Right: computation of a . O c P P and O c P P . The former is ℓ/
2, since it is the base angle of an isosceles triangle withbasis 1 and legs 1 − r (see also Fig. 7). The latter can be determined by difference and equals h − a/ b > a , we have to study the function G : h a ( h ) / − h − ℓ/ h ∈ (0 , ℓ ) that are the possible values for the parameter h . Its derivative is given by G ′ ( h ) = 3(1 − r ) cos h q − (1 − r ) sin h − . Since the function c c/ p − (1 − r ) (1 − c ) is increasing (its derivative has the sign of1 − (1 − r ) ), we see that G ′ h ) ≥ G ′ (0) = 3(1 − r ) − > r ≤ /
2. Thus, G isincreasing. Finally G ( ℓ ) = 0 because arcsin((1 − r ) sin ℓ ) = arcsin(sin( ℓ/ ℓ/
2, therefore G ( h ) < ⇔ b ( h ) > a ( h ) for h < ℓ . (cid:3) By definition and in view of the last proposition, the parameter associated to a cluster isbetween 0 and ℓ ( r ). Another constraint comes from the fact that the perimeter of Reuleauxpolygons of width 1 is π : given a rigid configuration of inradius r , 2 N + 1 sides, and m clustersof parameters h , . . . h m ∈ (0 , ℓ ( r )), there holds m X i =1 [ a ( r, h i ) + 2 b ( r, h i )] + (2 N + 1 − m ) ℓ ( r ) = π, ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 11 where a and b are the functions defined above. Recalling the relation (3.4), we get2 m X i =1 h i + (2 N + 1 − m ) ℓ ( r ) = π. (3.5) Remark 3.9.
The constraint (3.5) can be written in a more general form, allowing the param-eters h i to take also the values 0 and ℓ ( r ). Indeed, as already noticed in Remark 3.4, extremalarcs can be seen as degenerate cases of clusters: when h = 0 the arc Γ k reduces to a pointwhereas the two opposite sides Γ k − and Γ k form a unique arc of length ℓ ( r ); when h = ℓ ( r ),the triple (Γ k − , Γ k , Γ k +1 ) is of extremal arcs. In both cases, the formulas above for a , b , andperimeter are still valid. Therefore, every rigid shape can be described in terms of a collectionof parameters h i , i = 1 , . . . , ˜ m , varying in the closed interval [0 , ℓ ( r )]. The necessary condition(3.5) reads ˜ m X i =1 [2 h i + ℓ ( r )] = π. (3.6)In the remaining part of the section, we define a family of rigid shapes { Ω( r ) } r , whose opti-mality for A ( r ) will be proven in the next section. Definition 3.10.
Let r ∈ [1 − / √ , / N ( r ) := (cid:22) π ℓ ( r ) − (cid:23) , (3.7)where ⌊·⌋ denotes the integral part. This is the inverse of the function which associates to r theunique N ∈ N such that r ∈ ( r N − , r N +1 ]. We define Ω( r ) as the regular (2 N ( r ) + 1)-gon if r = r N ( r )+1 , and as the unique rigid shape with 2 N ( r ) + 1 sides and only one cluster. In thislast case, the parameter h associated to the cluster is uniquely determined by r , thus we maydenote it by h ( r ): in view of (3.5), it reads h ( r ) := π − (2 N ( r ) − ℓ ( r )2 . (3.8)4. Proof of Theorem 1.2 and Proposition 1.3
This section is devoted to the proofs of the main results. As announced in the Introduction,the key point is a density argument. As a first step we address the problem A N , N ∈ N , of areaminimization restricted to the class of Reuleaux polygons with at most 2 N + 1 sides. We willprove the following. Theorem 4.1.
The area minimization problem restricted to the family of Reuleaux polygonswith at most N + 1 sides, N ∈ N , has the following solution: A N ( r ) = (cid:26) + ∞ if N < N ( r ) | Ω( r ) | if N ≥ N ( r ) , (4.1) where N ( r ) and Ω( r ) are the function and the shape introduced in Definition 3.10. Moreover,when N ≥ N ( r ) , the minimizer is unique (up to rigid motion). Note that the equality A N ( r ) = + ∞ appearing in (4.1) is formal: we will show that the classof admissible shapes is empty for N < N ( r ).In order to prove Theorem 4.1, we need to compute the area of a rigid shape. To this aim,we split a shape with M sides into M subdomains, by connecting with straight segments theorigin to the vertexes. As previously, the origin is put at the center of the minimal annulus. Theelements of this partition can be regrouped as triples of subdomains associated to clusters andsubdomains associated to extremal arcs. In the next lemma we provide a formula for the areasof these subdomains. Lemma 4.2.
Let r be the inradius. Then the area of a triple of subdomains associated to acluster of parameter h is F ( r, h ) := (1 − r ) sin h cos h + a − sin a − r ) (cid:0) cos( a/ − (1 − r ) cos h (cid:1) sin( h + ℓ )+ b − sin b, (4.2) where ℓ = ℓ ( r ) , a = a ( r, h ) and b = b ( r, h ) are the functions introduced in (3.2) , (3.3) , and (3.4) ,respectively. Proof.
The area F ( r, h ) is the sum of two terms: F ( r, h ) = | A | + 2 | B | , where A is the subdomainwith boundary arc of length a and B is one of the two subdomains with boundary arc of length b (which clearly have the same area). Each of them can be furtherer decomposed as a triangleof the form OP j P j +2 and a portion of disk. For the subdomain A , the triangle is isosceles: thetwo sides which meet at O have length 1 − r and meet with an angle of 2 h , therefore the area is(1 − r ) sin h cos h. The area of the remaining part can be computed by difference, as the area of the circular sectorwith vertexes P j P j +1 P j +2 and the triangle with the same vertexes. The result is a − sin a . Let us now consider B . The triangle in B (not isosceles) has the following structure: the twosides which meet at O have length 1 − r and cos( a/ − (1 − r ) cos h , respectively, and form anangle of amplitude ℓ + h ; therefore its area is12 (1 − r ) (cid:0) cos( a/ − (1 − r ) cos h (cid:1) sin( h + ℓ ) . As already done for A , it is immediate to check that the remaining part in B has area b − sin b . By summing up the contributions we find (4.2). (cid:3)
Remark 4.3.
Notice that the formula above is valid also for h = 0 or ℓ , with the appropriateinterpretation. As already noticed in Remarks 3.4 and 3.9, when h = 0, the cluster reduces to asingle extremal arc. The formula above at 0 gives F ( r,
0) = (1 − r ) sin ℓ cos ℓ + ℓ − sin ℓ h = ℓ we have three extremal arcs, which is inaccordance to F ( r, ℓ ( r )) = 3 F ( r, . The properties of F are summarized in the following. Proposition 4.4.
The first and second derivatives of F with respect to the second variable aregiven by ∂F ( r, h ) ∂h =1 + 2(1 − r ) cos(2 h ) + 2(1 − r ) cos h cos( a/ (cid:0) − r ) sin h − (cid:1) , (4.3) ∂ F ( r, h ) ∂h = − − r ) sin(2 h ) + 2(1 − r ) sin h cos h cos ( a/
2) +2(1 − r ) sin h cos( a/ (cid:0) − − r ) sin h + 3(1 − r ) cos h (cid:1) , (4.4) where a = a ( r, h ) is the function introduced in (3.3) . In particular, ∂ F ( r, h ) ∂h < for any h ∈ [0 , ℓ ( r )] , ℓ ( r ) being the function introduced in (3.2) .Proof. Throughout the proof r is fixed, therefore we omit the dependence on it. In particular, F , a , and b , introduced in (4.2), (3.3), (3.4), respectively, will be regarded as functions of thesole variable h , and their derivatives will be denoted simply by a prime.We will use the following formulae, which can be deduced from tan( ℓ/
2) = p − r ) − ℓ/
2) = 12(1 − r ) , sin( ℓ/
2) = p − r ) − − r ) , (4.5)and cos ℓ = 12(1 − r ) − , sin ℓ = p − r ) − − r ) . (4.6) ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 13
From the definition of a and b , we have a ′ = 2(1 − r ) cos h cos( a/ , b ′ = 1 − (1 − r ) cos h cos( a/ . (4.7)Differentiating F and using (4.7) yields: F ′ ( h ) = (1 − r ) cos(2 h ) + (1 − r )(1 − cos a ) cos h cos( a/ + (cid:16) (1 − r ) sin h − (1 − r ) sin h cos h cos( a/ (cid:17) (1 − r ) sin( h + ℓ )+(cos( a/ − (1 − r ) cos h )(1 − r ) cos( h + ℓ ) + (1 − cos b ) (cid:16) − (1 − r ) cos h cos( a/ (cid:17) . Using (3.3), (3.4), and (4.5), we obtaincos a = 1 − − r ) sin h, cos( a/
2) = 1 − (1 − r ) sin h cos( a/ , and cos b = cos h cos (cid:0) ℓ − a (cid:1) − sin h sin (cid:0) ℓ − a (cid:1) = cos( a/ − r ) h cos h − sin h p − r ) − i + sin h h cos h p − r ) − h i . These computations allow to simplify the expression above of F ′ and to get (4.3).Differentiating one more time (4.3) we get F ′′ ( h ) = − − r ) cos h sin h + 2(1 − r ) sin h cos h (cid:2) − r ) sin h − (cid:3) / cos ( a/ − r ) sin h (cid:2) − − r ) sin h + 4(1 − r ) cos h (cid:3) / cos( a/ . Finally, writing 2(1 − r ) sin h − − r ) sin h − cos ( a/
2) and reordering the terms, wearrive at (4.4).Let us now prove that F ′′ ( h ) < h ∈ [0 , ℓ ]. To this aim, we write the second derivativeas F ′′ = 2(1 − r ) sin h ( A + B ), with A ( r, h ) := −
19 (1 − r ) cos h + (1 − r ) sin h cos h cos ( a/ B ( r, h ) := −
359 (1 − r ) cos h + 3(1 − r ) cos h − − r ) sin h + 1cos( a/ . If we prove that A and B are negative, we are done.Since h a ( h ) is increasing, both terms in A are increasing. Therefore A ( r, h ) ≤ A ( r, ℓ ).Using (4.5) and (4.6) we get A ( r, ℓ ) = −
19 (1 − r ) (cid:20) − r ) − (cid:21) + 8(1 − r ) − r ) − − r ) (cid:20) − r ) − (cid:21) . This leads to look at the sign of the polynomial P ( x ) = − x + 3 x − , with x := 1 − r . Sincethe roots of P are 1 / √ p / P is negative in [ , √ ], we conclude that A ( r, h ) ≤ B ( r, h ). It has the same sign of B ( r, h ) = −
359 (1 − r ) cos h cos( a/
2) + 1 + (1 − r ) h ) + 1) . Now, comparing their sin, it is immediate that, for any h , a/ ≤ (1 − r ) h . Therefore, B ( r, h ) ≤ B ( r, h ) = −
359 (1 − r ) cos h cos((1 − r ) h ) + 1 + (1 − r ) h ) + 1) . We now compute the three first derivatives of h B ( r, h ). It comes, after linearisation d B dh = (1 − r ) (cid:20) − (cid:0) r sin( rh ) + (2 − r ) sin((2 − r ) h ) (cid:1) + 20(1 − r ) sin(2 h ) (cid:21) . Using sin( rh ) ≤ rh , sin((2 − r ) h ) ≤ (2 − r ) h and sin(2 h ) ≥ h/π we get d B dh ≥ (1 − r ) h (cid:20) − (cid:0) r + (2 − r ) (cid:1) + 80 π (1 − r ) (cid:21) . Since r + (2 − r ) ≤ / − r ≥ / d B dh ≥ dB dh is convex in h , moreover it vanishes at 0. Thus, dB dh is either always positive or always negative or negativeand then positive (and this is actually the case). In any case, we see that B ( r, h ) ≤ max ( B ( r, , B ( r, ℓ )) . Now we see that B ( r,
0) = 3(1 − r ) − (1 − r ) + 1 ≤ − < B ( r, ℓ ). For that purpose, we claim the following:cos((1 − r ) ℓ ) ≥ −
125 (1 − r ) = 125 r − . (4.8)Recalling the relation (3.2) between ℓ and r , the validity of (4.8) is related to the positivity ofthe auxiliary function ψ ( r ) := arccos (cid:18) r − (cid:19) − − r ) arctan (cid:16)p − r ) − (cid:17) . The second derivative of ψ reads ψ ′′ ( r ) = − r − − (12 r − ) / + 2(1 − r )(4(1 − r ) − / and is negative in [1 − / √ , / ψ is concave and ψ ( r ) ≥ min( ψ (1 − / √ , ψ (1 / ≥ . This proves the claim.We insert the estimate (4.8) in B to get (we still use x = 1 − r ): B ( r, ℓ ) ≤ − x (cid:18) x − (cid:19) (11 − x ) + 1 + x (cid:18) x − (cid:19) − x − x + 1 ! . This leads to consider the polynomial Q ( x ) = − x + 779 x + 23 x − x + 54 . This polynomial is negative in [1 / , / √ B is negative too. (cid:3) Proof of Theorem 4.1.
We begin by noticing that if
N < N ( r ), then the class of admissibleshapes is empty: assume by contradiction that there exists a Reuleaux polygon contained intothe annulus B − r (0) \ B r (0) with M < N ( r ) + 1 sides. Each arc of the boundary has length atmost ℓ ( r ), therefore, imposing that the perimeter is π and recalling the definition (3.7) of N ( r ),we get π ≤ M ℓ ( r ) < (2 N ( r ) + 1) ℓ ( r ) = (cid:18) (cid:22) π ℓ ( r ) − (cid:23) + 1 (cid:19) ℓ ( r ) ≤ π, which is absurd.Let now N ≥ N ( r ). The proof is divided into four steps. Step 1.
By Definition 3.1, any shape that is not rigid can be modified, through an admissibleBlaschke deformation, to decrease the area. Thus, it remains to minimize the area among rigidshapes. We have seen in Proposition 3.5 that these rigid shapes are composed of extremal arcsand clusters.
Step 2.
Let us write an area formula for a rigid shape Ω. Connecting with straight segmentsthe origin to the vertexes, we split Ω into subdomains, which can be regrouped as triples ofsubdomains associated to clusters and subdomains associated to extremal arcs. According tothe notation used in Remark 3.9, all the subdomains can be regarded associated to clusters,allowing the parameters h i to vary in the closed interval [0 , ℓ ( r )], i = 1 , . . . , ˜ m , for a suitable˜ m ∈ N . In view of Lemma 4.2 and Remark 4.3, we infer that the total area is | Ω | = ˜ m X i =1 F ( r, h i ) . Notice that the area does not explicitly depend on the relative position of the clusters (thisdependence is enclosed into the relation among the lengths). The necessary condition (3.6) gives
ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 15 a restriction on the possible values of ˜ m : since every h i is between 0 and ℓ ( r ), we infer that2 h i + ℓ ( r ) ∈ [ ℓ ( r ) , ℓ ( r )], so that, summing over i from 1 to ˜ m , we get π ℓ ( r ) ≤ ˜ m ≤ (cid:22) πℓ ( r ) (cid:23) . (4.9)In particular, this implies that the number of sides of a rigid configuration cannot be arbitrarilylarge, but it is bounded by a quantity depending only on r . We infer that the sequence of minima {A N ( r ) } N ≥ N ( r ) is constant after a finite number of values (depending on r ). A priori, the firstterms of the sequence could be different. In the next step we show that, actually, the sequenceis constant in N . Step 3.
Let us optimize the area when ˜ m is fixed. In view of the previous step, we are led tominimize the function F ( h , . . . , h ˜ m ) := ˜ m X i =1 F ( r, h i ) , over the set C := ( ( h , . . . , h ˜ m ) ∈ [0 , ℓ ( r )] ˜ m , ˜ m X i =1 [2 h i + ℓ ( r )] = π ) . In that way, we transform a geometric problem into an analytic one which might have solutionsthat do not correspond to real geometric shapes. It turns out, as we will see below, that theminimizer is unique and actually corresponds to a real body of constant width.The set of constraints is the intersection of an hypercube and an hyperplane. In view ofProposition 4.4, the function F is strictly concave, therefore it attains a minimum on extremalpoints of C . The extremal points of C lie on the edges of the hypercube, namely (up to relabeling) h ∈ [0 , ℓ ] and h , . . . , h ˜ m ∈ { , ℓ } . Without loss of generality, we may label the h i s in such away that h , . . . , h q = 0 and h q +1 , . . . , h ˜ m = ℓ . We claim the following facts:(i) for ˜ m fixed, the extremal point of C is unique;(ii) for a fixed r , the minimum of F does not depend on ˜ m in the range (4.9).The case in which r is the inradius of some regular Reuleaux polygon is trivial: in view of (3.6),the parameter h has to belong to { , ℓ ( r ) } and, again by (3.6), no matter how the sides areregrouped (one by one when h = 0, three by three when h = ℓ ( r )), they are necessarily 2 N ( r )+1,where N ( r ) is the number introduced in (3.7).In all the other cases, h lies necessary between 0 and ℓ ( r ), strictly. A first consequence isthat the number of sides is 3 + ( q −
1) + 3( ˜ m − q ). Since it is odd, we infer that ˜ m is odd, too.In view of (3.6), q is given by q = 32 ˜ m − π ℓ ( r ) + h ℓ ( r ) . More precisely, taking into account that h /ℓ ( r ) ∈ (0 , q = q ( ˜ m ) := 1 + (cid:22)
32 ˜ m − π ℓ ( r ) (cid:23) . Using again (3.6), we infer that h is given by h = ℓ ( r )(1 − δ ) , with δ := 32 ˜ m − π ℓ ( r ) − (cid:22)
32 ˜ m − π ℓ ( r ) (cid:23) ∈ (0 , . All in all, once fixed ˜ m , h and q are determined. This concludes the proof of (i).Notice that if we replace ˜ m by ˜ m + 2 (as already noticed ˜ m has to be odd), the value of δ does not change. This allows us to write h , without the dependence on ˜ m . Therefore, in orderto prove (ii), it is enough to show that the number of sides of length ℓ ( r ) does not depend on ˜ m : q ( ˜ m ) − m − q ( ˜ m )) = 3 ˜ m − q ( ˜ m ) − m − (cid:22)
32 ˜ m − π ℓ ( r ) (cid:23) −
3= 3( ˜ m − − (cid:22)
32 ( ˜ m − − (cid:18) π ℓ ( r ) − (cid:19)(cid:23) = 2 (cid:18)(cid:22) π ℓ ( r ) − (cid:23) + 1 (cid:19) . Here we have used that ˜ m − ⌊ k − x ⌋ = k − ⌊ x ⌋ −
1, true forevery k ∈ N and every 0 < x < k , x / ∈ N . This proves (ii). Step 4.
In view of the previous step, we immediately get that the optimal shape associated tothe inradius r of a regular Reuleaux polygon, is the Reuleaux polygon itself, for every N ≥ N ( r ).When r is not the inradius of a regular Reuleaux polygon, we have shown that, for every N ≥ N ( r ), the optimal configuration has a unique cluster and have all the other sides of length ℓ ( r ). As already underlined in Definition 3.10, these properties characterize the set Ω( r ), andthe proof of the theorem is concluded. Note that 2 ⌊ π/ (2 ℓ ( r )) − / ⌋ + 2 (i.e. the number ofsides of length ℓ ( r ) found in Step 3) is equal to N ( r ) −
2, implying that the total number of sidesof the optimal shape is 2 N ( r ) + 1, as expected. (cid:3) We are now in a position to prove the main results, about the characterization of minimizersand the continuity of minima and minimizers with respect to r . Proof of Theorem 1.2.
In view of the density of the Reuleaux polygons in the class of constantwidth sets see [3] or [4], we infer that A ( r ) = inf N ∈ N A N ( r ) . In view of Theorem 4.1, we infer that the sequence A N ( r ) is finite and constant after N ( r ), sothat A ( r ) = inf N A N ( r ) = | Ω( r ) | . The other statements follow from the characterization of Ω( r )(see Definition 3.10 and Proposition 3.8). (cid:3) Proof of Proposition 1.3.
In view of Theorem 1.2, its proof, and Definition 3.10, A ( r ) can becomputed by dividing the optimal shape Ω( r ) into 2 N ( r ) + 1 subdomains, obtained by joiningwith segments the vertexes with the origin. The partition is made of subdomains associated toextremal arcs of length ℓ ( r ) and (possibly) to one triple associated to the cluster of parameter h ( r ). According to (4.2), the former have all area F ( r, F ( r, h ( r )). When r = r N +1 the partition is regular and A ( r N +1 ) = (2 N + 1) F ( r N +1 , . In all the other cases, namely when r ∈ ( r N − , r N +1 ), we have A ( r ) = (2 N − F ( r,
0) + F ( r, h ( r )) . In the open interval ( r N − , r N +1 ) the functions F ( r, h ( r ), and F ( r, h ( r )) are continuous,therefore A ( r ) is continuous too. In the limit as r ց r N − , we have h ( r ) →
0, so thatlim r ց r N − A ( r ) = (2 N − F ( r N − ,
0) + F ( r N − ,
0) = A ( r N − );similarly, when r ր r N +1 , we have h ( r ) → ℓ ( r N +1 ) and F ( r, ℓ ( r )) → F ( r N +1 , r ր r N +1 A ( r ) = (2 N − F ( r N +1 ,
0) + 3 F ( r N +1 ,
0) = (2 N + 1) F ( r N +1 ,
0) = A ( r N +1 ) . Therefore, A is continuous in each closed interval [ r N − , r N +1 ]. This concludes the proof of thecontinuity of A .Let us now consider the optimal shapes. We choose the following orientation: for regularReuleaux polygons, we take one of the vertexes aligned vertically with the origin, above it; inall the other cases, we choose the point P k of the cluster (see Definition 3.3) aligned verticallywith the origin, below it (see also Fig. 1). By construction, the position of the vertexes variescontinuously with respect to r , so that the optimal shapes vary continuously with respect to theHausdorff convergence. (cid:3) Acknowledgements : The authors want to thank G´erard Philippin for stimulating discus-sions. This work was partially supported by the project ANR-18-CE40-0013 SHAPO financedby the French Agence Nationale de la Recherche (ANR). IL acknowledges the Dipartimento diMatematica - Universit`a di Pisa for the hospitality.
ODY OF CONSTANT WIDTH WITH MINIMAL AREA IN A GIVEN ANNULUS 17
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Antoine
Henrot , Universit´e de Lorraine CNRS, IECL, F-54000 Nancy, France email: [email protected]
Ilaria
Lucardesi , Universit´e de Lorraine CNRS, IECL, F-54000 Nancy, France email: [email protected] hP k P k − P k +2 P k +1 P k −−