Envelopes of α -sections
aa r X i v : . [ m a t h . M G ] S e p Envelopes of α -sections N. Chevallier, A. Fruchard, and C. Vˆılcu
Abstract
Let K be a planar convex body af area | K | , and take 0 < α <
1. An α -section of K is a line cutting K into two parts, one of which has area α | K | . This article presents asystematic study of the envelope of α -sections and its dependence on α . Several openquestions are asked, one of them in relation to a problem of fair partitioning. Keywords:
Convex body, alpha-section, envelope, floating body, fair partitioning.
MSC Classification: 52A10, 52A38, 51M25, 51M04
In this paper, unless explicitly stated otherwise, K denotes a convex body in the Euclideanplane E ; i.e., a compact convex subset of E with nonempty interior. Let ∂K denote theboundary of K and | K | its area. Given α ∈ ]0 , α -section of K is an oriented line∆ ⊂ E cutting K in two parts, one to the right, denoted by K − , of area | K − | = α | K | ,and the other to the left, K + , of area | K + | = (1 − α ) | K | ; here K ± are compact sets, thus K + ∩ K − = ∆ ∩ K .Denote by K α the intersection of all K + and call it the α -core of K ; denote by m α the envelope of all α -sections of K .The purpose of this article is to study m α and its relation to the α -core. We referto Sections 6-7 for formal statements, and give in this introductory section an informalpresentation. Since K α is empty for α > , we will often implicitly assume α ≤ whendealing with α -cores. The situation depends essentially upon whether K is centrallysymmetric (we will say ‘symmetric’ in the sequel, for short) or not. Precisely, we provethe following statements.If K is symmetric then one has m α = ∂K α for all α ∈ (cid:3) , (cid:2) . Moreover, we have thefollowing equivalence: the envelope m α is of class C for all α ∈ (cid:3) , (cid:2) if and only if K isstrictly convex.If K is non-symmetric then we cannot have m α = ∂K α for all α ∈ (cid:3) , (cid:2) , because m α exists for all α , whereas K α is empty for α close enough to . More precisely, there existsa critical value α B ∈ (cid:2) , (cid:2) such that for all α ∈ ]0 , α B ] we have m α = ∂K α , and for all α ∈ (cid:3) α B , (cid:2) we have m α ) ∂K α .The case α B = 0 can occur, e.g. , if there exists a triangle containing K with an edgeentirely contained in ∂K . We also prove that m α is never of class C for α ∈ (cid:3) α B , (cid:2) , andthat m α is of class C for all α ∈ ]0 , α B [ if and only if ∂K does not contain two parallelsegments. As a by-product, we obtain the following characterization: A convex body K is non-symmetric if and only if there exists a triangle containing more than half of K (inarea), with one side entirely in K and the two others disjoint from the interior of K . oncerning the α -core, we prove that there is another critical value, α K ∈ (cid:2) , (cid:3) , suchthat if 0 < α < α K then K α is strictly convex with nonempty interior, if α = α K then K α is reduced to one point, and if α K < α < K α is empty. We emphasize that,when K α is a point, this point is not necessarily the mass center of K , see Section 8.9.The value α K = occurs if and only if K is symmetric and the value α K = occurs ifand only if K is a triangle.A similar study, for secants between parallel supporting lines to K , whose distances tothe corresponding lines make a ratio of α/ (1 − α ), is the subject of (ir)reducibility theory of convex bodies. There too, the envelope of those secants is sometimes different from theintersection of the half-planes they are defining; and there exists a ratio, called critical,for which the later object is reduced to a point. See for example [14, 18, 32].Our paper is closely related to previous works about slicing convex bodies, outerbilliards (also called dual billiard ), floating bodies, and fair (or equi-) partitioning. It isalso related to continuous families of curves in the sense of Gr¨unbaum, see Section 8.8and the references therein. There is a vast literature on these subjects; we refer in thefollowing to very few articles, and briefly present even fewer, that we find particularlyrelevant for our study. Further references can be found in those papers.Generalizing previous results on common tangents and common transversals to familiesof convex bodies [2], [6], J. Kincses [17] showed that, for any well-separated family ofstrictly convex bodies, the space of α -sections is diffeomorphic to S d − k .A billiard table is a planar strictly convex body K . Choose a starting point x outsidethe table and one of the two tangents through x to K , say the right one, denoted by D ;the image T ( x ) of x by the billiard map T is the point symmetric of x with respect to thetangency point D ∩ ∂K . A caustic of the billiard is an invariant curve (an invariant torusin the terminology of the KAM theory). The link between outer billiards and α -sectionsis the following: If K is the envelope of α -sections of a convex set bounded by a curve L ,for some α , then L is a caustic for the outer billiard of table K , cf. e.g. [11].Outer billiards have been considered by several authors, such as J. Moser [22], V. F. La-zutkin [19], E. Gutkin and A. Katok [13], D. Fuchs and S. Tabachnikov [11], S. Tabach-nikov [29]. Therefore, it is natural that several authors considered envelopes of α -sectionsin the framework of outer billiard, see e.g. , Lecture 11 of the book of D. Fuchs andS. Tabachnikov [11], and references therein. Besides those studies, the envelope of α -sections seems to be scarcely studied.With our notation, the set K [ α ] bounded by m α was called floating body of K andits study goes back to C. Dupin, see [9, 33]. On the other hand, our “ α -core” K α wasintroduced by C. Sch¨utt and E. Werner [25] and studied in a series of papers [25, 26,28, 33]. under the name of convex floating body . For convex bodies K in R d and for α small enough, they gave estimates for vol n ( K ) − vol n ( K [ α ] ) and for vol n ( K ) − vol n ( K α ),in relation to the affine surface area and to polygonal approximations. M. Meyer andS. Reisner [21] proved in arbitrary dimension that K is symmetric if and only if m α = ∂K α for any α ∈ (cid:3) , (cid:2) . They also prove that m α is smooth if K is strictly convex. A. Stancu[28] considered convex bodies K ⊂ R d with boundary of class C ≥ , and proved that thereexists δ K > K δ is homothetic to K , for some δ < δ K , if and only K is anellipsoid.The terminology of “(convex) floating body” is very suggestive for the floating theoryin mathematical physics. For our study however, considering its close connections to α -sections and fair partitioning, it seems more natural to use the term of “ α -core”.Some of our results, essentially Proposition 4.1, are known and already published. Inthat case we mention references after the statement. For the sake of self-containedness,however, we will provide complete proofs.We end the paper with several miscellaneous results and open questions. Our main onjecture is as follows: If K ⊆ L are two convex bodies and α ∈ (cid:3) , (cid:2) , then there existsan α -section of L which either does not cross the interior of K , or is a β -section of K forsome β ≤ α . The conjecture has been recently proven in the case of planar convex bodies,see [10].There are two reasons which motivated us to undertake this systematic study of α -sections and their envelopes. First, we found only one reference which focuses especiallyon the envelope of α -sections, Lecture 11 of the nice book [11], which is a simplifiedapproach. Although the results contained in the present article seem natural, and theproofs use only elementary tools and are most of the time simple, we hope that our workwill be helpful in clarifying the things.Our second motivation for studying α -sections and their envelope was a problem of fairpartitioning of a pizza. What we call a pizza is a pair of planar convex bodies K ⊆ L , where L represents the dough and K the topping of the pizza. The problem of fair partitioningof convex bodies in n pieces is a widely studied topic, see e.g. [1, 3, 4, 5, 15, 16, 24, 27].Nevertheless, to our knowledge, our way of cutting has never been considered: We usea succession of double operations: a cut by a full straight line, followed by a Euclideanmove of one of the resulting pieces; then we repeat the procedure. The final partition issaid to be fair if each resulting slice has the same amount of K and the same amount of L . The result of [10] is the following: Given an integer n ≥
2, there exists a fair partitionof any pizza (
K, L ) into n parts if and only if n is even. The notation S stands for the standard unit circle, S := R / (2 π Z ), endowed with itsusual metric d ( θ, θ ′ ) = min {| τ − τ ′ | ; τ ∈ θ, τ ′ ∈ θ ′ } . On S we use the notation: θ ≤ θ ′ if there exist τ ∈ θ, τ ′ ∈ θ ′ such that τ ≤ τ ′ < τ + π ; it is not an order because it is nottransitive.Given θ ∈ S , let ~u ( θ ) denote the unit vector of direction θ , ~u ( θ ) = (cos θ, sin θ ).For convenience, we add arrows ~ on vectors. Unless explicitly specified otherwise, allderivatives will be with respect to θ , hence e.g. , ~u ′ ( θ ) = d~udθ ( θ ) = ( − sin θ, cos θ ) is the unitvector orthogonal to ~u ( θ ) such that the frame (cid:0) ~u ( θ ) , ~u ′ ( θ ) (cid:1) is counterclockwise.Given an oriented straight line ∆ in the plane, ∆ + denotes the closed half-plane onthe left bounded by ∆, and ∆ − is the closed half-plane on the right. We identify orientedstraight lines with points of the cylinder C = S × R , associating each pair ( θ, t ) ∈ C tothe line oriented by ~u ( θ ) and passing at the signed distance t from the origin. In otherwords, the half-plane ∆ + is given by ∆ + = { x ∈ R ; h x, ~u ′ ( θ ) i ≥ t } . We endow C withthe natural distance d (cid:0) ( θ, t ) , ( θ ′ , t ′ ) (cid:1) = (cid:0) d ( θ, θ ′ ) + | t − t ′ | ) (cid:1) / .Given α ∈ ]0 , α -section of K is an oriented line ∆ such that | ∆ − ∩ K | = α | K | .For all α ∈ ]0 ,
1[ and all θ ∈ [0 , π [, there exists a unique α -section of K of direction θ ;it will be denoted by ∆( α, θ ). This defines a continuous function ∆ : ]0 , × S → C . Weobviously have the symmetry∆ ± (1 − α, θ ) = ∆ ∓ ( α, θ + π ) . (1)For 0 < α ≤ , we call α -core of K , and denote by K α , the intersection of all lefthalf-planes bounded by α -sections: K α = \ θ ∈ S ∆ + ( α, θ ) . It is a compact convex subset of the plane, possibly reduced to one point or empty. et ∆ = ∆( α, θ ) be an α -section of K of direction ~u = ~u ( θ ), and let b, c denote theendpoints of the chord ∆ ∩ K , with ~bc having the orientation of ~u ( i.e. , the scalar product h ~bc, ~u i is positive), see Figure 1. ✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✥✿✥✥✿ ~u ❉❉❉❖ ~u ′ ❛❛❛❛❛q r ~b ′ b β r m ✄✄✄✄✗ r ~c ′ c γK ∂K ∆ ∆ − ∆ + Figure 1:
Some notationLet m = ( b + c ) denote the midpoint of the chord bc . Let h denote the half-length of bc ; hence we have b = m − h~u and c = m + h~u . The functions b, c, m and h are continuouswith respect to α and θ . As we will see in the proof of Proposition 4.1, they are also left-and right-differentiable with respect to θ at each θ ∈ S , e.g. , the following limit exists(with the convention θ → θ − for θ → θ , θ < θ ) ~b ′ l ( θ ) = lim θ → θ − θ − θ (cid:0) b ( α, θ ) − b ( α, θ ) (cid:1) , and similarly for ~b ′ r , ~c l ′ , ~c r ′ . They are also left- and right-differentiable with respect to α ,but we will not use this fact.We say that b is a regular point of ∂K ( regular for short) if there is a unique supportingline to K at b ( i.e. , ~b ′ l = ~b ′ r ); otherwise we call b a corner point of ∂K (a corner for short).If b is regular then β = ( d ~b ′ , ~u ) ∈ ]0 , π [ denotes the angle between the tangent to ∂K at b and ∆. If b is a corner then β l = ( d ~b ′ l , ~u ), resp. β r = ( d ~b ′ r , ~u ), is the angle between theleft-tangent, resp. right-tangent, to ∂K at b and ∆. Similarly, let γ = ( d ~u, ~c ′ ) ∈ ]0 , π [ (if c is regular), resp. γ l , γ r (if c is a corner), be the angle between ∆ and the tangent, resp.left-tangent, right-tangent, to ∂K at c , see Figure 1.For a fixed α , the values of θ such that b ( α, θ ) or c ( α, θ ) (or both) is a corner will becalled singular ; those for which both b and c are regular will be called regular .Observe that we always have β r ≤ β l and γ l ≤ γ r , with equality if and only if b , resp. c , is regular. Also observe the following fact. b and c admit parallel supporting lines if and only if β r + γ l ≤ π ≤ β l + γ r . (2)Finally, observe that the angle between ~b ′ , (resp. ~b ′ l , ~b ′ r ) and the axis of abscissae is equalto θ − β ( α, θ ) (resp. θ − β l ( α, θ ), θ − β r ( α, θ )). Since these angles are increasing andintertwining functions of θ , we have the following statement.If θ < θ ′ then θ − β l ( α, θ ) ≤ θ − β r ( α, θ ) ≤ θ ′ − β l ( α, θ ′ ) ≤ θ ′ − β r ( α, θ ′ ) . It follows thatlim θ → θ +0 β l ( α, θ ) = lim θ → θ +0 β r ( α, θ ) = β r ( α, θ ) ≤ β l ( α, θ ) = lim θ → θ − β l ( α, θ ) = lim θ → θ − β r ( α, θ ) , (3) nd similarlylim θ → θ − γ l ( α, θ ) = lim θ → θ − γ r ( α, θ ) = γ l ( α, θ ) ≤ γ r ( α, θ ) = lim θ → θ +0 γ l ( α, θ ) = lim θ → θ +0 γ r ( α, θ ) . (4)In a same way we haveIf α < α ′ then β r ( α, θ ) ≤ β l ( α, θ ) ≤ β r ( α ′ , θ ) ≤ β l ( α ′ , θ )and γ l ( α, θ ) ≤ γ r ( α, θ ) ≤ γ l ( α ′ , θ ) ≤ γ r ( α ′ , θ ) . (5)As a consequence,lim α → α − β l ( α, θ ) = lim α → α − β r ( α, θ ) = β r ( α , θ ) ≤ β l ( α , θ ) = lim α → α +0 β l ( α, θ ) = lim α → α +0 β r ( α, θ ) , lim α → α − γ l ( α, θ ) = lim α → α − γ r ( α, θ ) = γ l ( α , θ ) ≤ γ r ( α , θ ) = lim α → α +0 γ l ( α, θ ) = lim α → α +0 γ r ( α, θ ) . All these elements b, c, m, h, β, γ can be considered as functions of both α and θ . Never-theless, as already said, all derivatives are with respect to θ .Let v = v ( α, θ ) be the scalar product v = h ~m ′ , ~u i ∈ R . If v has a discontinuity, then v l and v r denote its corresponding left and right limits. As we will see, ~m ′ is always collinearto ~u , hence v is the “signed norm” of ~m ′ . We will also see that v has discontinuities onlyif b or c (or both) is a corner. Since we chose θ as parameter, we will also see that v is thesigned radius of curvature of the curve m , but we prefer to refer to it as the velocity ofthe current point m of the envelope. The symmetry (1) gives m ( α, θ + π ) = m (1 − α, θ ),hence ~m ′ ( α, θ + π ) = ~m ′ (1 − α, θ ). Since ~u ( θ + π ) = − ~u ( θ ), we obtain v l ( α, θ + π ) = − v l (1 − α, θ ) and v r ( α, θ + π ) = − v r (1 − α, θ ) . (6)Our last notation is V for the segment of endpoints v l and v r ; i.e. , V = [ v l , v r ] if v l ≤ v r , V = [ v r , v l ] otherwise. Formulae (6) yield V ( α, θ + π ) = − V (1 − α, θ ) . (7) Before carrying on α -sections, we would like to digress for a moment in a more generalframework. The notion and results of this section are elementary and probably alreadyknown, but we did not find any reference in the literature. They can be considered asexercises in a graduate course on planar curves.Recall that a ruled function (or regulated function ) R : S → R is the uniform limitof piecewise constant functions. It is equivalent to saying that R admits a left- and aright-limit, denoted below by R l and R r , at each point of S , see e.g. , [8]. Recall also ournotation ~u ( θ ) = (cos θ, sin θ ). Definition 3.1 . (a) A tour is a planar curve parametrized by its tangent. Moreprecisely, we call m : S → R a tour if m is continuous, has a left- and a right-derivativeat each point of S , and if there exists a ruled function R : S → R such that ~m ′ l ( θ ) = R l ( θ ) ~u ( θ ) and ~m ′ r ( θ ) = R r ( θ ) ~u ( θ ).(b) The core K = K ( m ) of a tour m is the intersection of all left half-planes delimitedby all tangents to m oriented by ~u , i.e. , K = T θ ∈ S D + ( θ ), where D ( θ ) = m ( θ ) + R ~u ( θ ).Tours are not necessarily simple curves. The case m ( θ ) = m ( θ + π ) gives rise to a doublehalf-tour . This is the case e.g. for the envelope of half-sections of a planar convex body. Some (double half-)tours: Left the astroid, right the deltoid. The astroid cannot be anenvelope of an α -section of a planar convex body; the deltoid is probably such a half-section, althoughwe ignore how to prove it. Proposition 3.2 . Let m be a tour, with associated ruled function R , and let m ∗ denoteits image in the plane: m ∗ = m ( S ) . (a) For each θ ∈ S , R l ( θ ) , resp. R r ( θ ) , is the signed left-, resp. right-radius of curvatureof m ∗ at m ( θ ) . (b) If R l and R r do not vanish on S , then m ∗ is a C submanifold of the plane. (c) Conversely, if there exist θ ≤ θ ≤ θ + π ∈ S such that the product R r ( θ ) R l ( θ ) is negative, then there exists θ ∈ [ θ , θ ] such that m ∗ is not of class C at m ( θ ) . (d) The core K of m is either empty, or a point, or a strictly convex body. (e) The boundary of the core, ∂K , is included in m ∗ . Moreover, ∂K and m ∗ coincide ifand only if the functions R l and R r are nonnegative on S .Remarks.
1. In the context of convex floating bodies, the statement (d) above is already known,also in arbirtrary dimension, see e.g.
Prop. 1 (iv) and (v) in [33] or Theorem 3 in [21].2. In the case where R vanishes without changing of sign, m ∗ may, or may not, be of class C . Proof . (a) Let s denote the curvilinear abscissa on m from some starting point, say m (0).We have s ( θ ) = Z θ R l ( τ ) dτ = Z θ R r ( τ ) dτ . Locally, if R l ( θ ) and R r ( θ ) are nonzero,then the tangent of m at m ( θ ) is ~u ( θ ) and we have (cid:16) d~uds (cid:17) l/r ( θ ) = d~udθ (cid:16) dθds (cid:17) l/r ( θ ) = 1 R l/r ( θ ) ~u ′ ( θ ) . (b) If R l and R r do not vanish, then s is a homeomorphism (with inverse homeomorphismdenoted θ for convenience) and m (cid:0) θ ( s ) (cid:1) = m (0) + Z s ~u (cid:0) θ ( t ) (cid:1) dt , with ~u and θ continuous, i.e., m ◦ θ is of class C .(c) We may assume, without loss of generality, that R r ( θ ) > R l ( θ ) <
0. If θ = θ then m ( θ ) is a cusp ( i.e. , a point with one half-tangent), hence m ∗ is not C at m ( θ ). Inthe sequel we assume θ < θ . Consider E = { θ ∈ [ θ , θ ] ; R r ( θ ) > } . Let θ = sup E .Since 0 > R l ( θ ) = lim θ → θ − R l ( θ ) = lim θ → θ − R r ( θ ), we have θ / ∈ E for θ < θ , θ close enough o θ . In the same manner, we have θ ∈ E if θ > θ , θ close enough to θ . Therefore, wehave θ < θ < θ .Now two cases may occur. If there exists θ > θ , such that R vanishes on the wholeinterval ] θ , θ [ , then m is constant on [ θ , θ ]. We assume θ ≤ θ maximal with thisproperty. By contradiction, if m ∗ is C at the point m ( θ ) = m ( θ ) then necessarily wehave θ = θ + π , a contradiction with θ < θ ≤ θ ≤ θ + π .In the other case, for all δ > θ ∈ ] θ , θ + δ [ such that R r ( θ ) <
0, andwe obtain that m ( θ ) is a cusp.Before the proofs of (d) and (e), we first establish two lemmas. Lemma 3.3 . The interior of the core K of m coincides with the intersection of all openhalf-planes Int (cid:0) D + ( θ ) (cid:1) , θ ∈ S : Int ( K ) = \ θ ∈ S Int (cid:0) D + ( θ ) (cid:1) . (8) Furthermore, we have ∂K ⊂ S θ ∈ S D ( θ ) .Proof . The inclusion ⊆ in (8) is evident: For each θ we have K ⊂ D + ( θ ), hence Int ( K ) ⊂ Int (cid:0) D + ( θ ) (cid:1) . Conversely, given x ∈ K , the map θ dist (cid:0) x, D ( θ ) (cid:1) is continuous. Since S is compact, if x is in the interior of D + ( θ ) for all θ ∈ S , then this map has a minimum ρ >
0, and the disc of center x and radius ρ is included in K . This proves (8).If x ∈ ∂K , then x is in every closed half-plane D + ( θ ) but not in every open one by (8),hence x has to be on (at least) one of the lines D ( θ ). Lemma 3.4 . If θ ∈ S and z are such that z ∈ ∂K ∩ D ( θ ) then z = m ( θ ) .Proof . For small ε = 0, positive or negative, we have m ( θ + ε ) − m ( θ ) = Z θ + εθ R ( θ ) ~u ( θ ) dθ. By boundedness of R l and R r , and by continuity of ~u , we deduce that there exists r equalto R l ( θ ) or R r ( θ ) such that m ( θ + ε ) − m ( θ ) = r~u ( θ ) ε + o ( ε ) . (9)It follows that D ( θ ) and D ( θ + ε ) cross at a distance O ( ε ) from m ( θ ), see Figure 3 below.If z were different from m ( θ ) then, for ε small, either negative or positive depending onthe relative positions of z and m ( θ ), we would have z ∈ Int (cid:0) D − ( θ + ε ) (cid:1) , hence z / ∈ K ,a contradiction. Now we return to the proof of Proposition 3.2. (d) By contradiction, assume that ∂K contains some segment [ x, y ] with x = y and take z ∈ ] x, y [ arbitrarily. By Lemma 3.3, there exists θ ∈ S such that z ∈ D ( θ ). Sinceboth x and y belong to K ⊂ D + ( θ ), the line D ( θ ) contains both x and y , i.e., θ isthe direction of ± → xy . By Lemma 3.4, we deduce that z = m ( θ ). Hence we proved that any z ∈ ] x, y [ coincides with z = m ( θ ), where θ is one of the directions ± → xy , which isimpossible.(e) The first assertion follows directly from Lemmas 3.3 and 3.4. For the second one,first we proceed by contradiction and we assume R l ( θ ) < R r ( θ ) < θ ∈ S , say R r ( θ ) <
0. Then by (9), for ε > m ( θ + ε ) = r r ✏✏✏✏✏✏✏✏✏✏✏✏✏✏✶ r m ( θ ) m ( θ + ε ) D ( θ + ε ) D ( θ ) z Figure 3:
Proof of Lemma 3.4 and Proposition 3.2 (d) m ( θ ) + εR r ( θ ) ~u ( θ ) + o ( ε ), hence m ( θ ) ∈ Int (cid:0) D − ( θ + ε ) (cid:1) , see Figure 3. It follows that m ( θ ) / ∈ K .Conversely, if R l and R r are nonnegative, take θ ∈ S arbitrarily. Then, for all θ ∈ [ θ , θ + π ], we have m ( θ ) = m ( θ ) + Z θθ ~m ′ ( τ ) dτ = m ( θ ) + Z θθ R l ( τ ) ~u ( τ ) dτ. Therefore we obtain h → m ( θ ) m ( θ ) , ~u ′ ( θ ) i = Z θθ R l ( τ ) h ~u ( τ ) , ~u ′ ( θ ) i dτ ≥ , since h ~u ( τ ) , ~u ′ ( θ ) i = sin( θ − θ ) ≥
0. In the same manner, we have for all θ ∈ [ θ − π, θ ], h → m ( θ ) m ( θ ) , ~u ′ ( θ ) i ≥
0, hence the whole curve m ∗ is in the half-plane D + ( θ ). Thisholds for all θ ∈ S , hence m ∗ ⊂ K . Finally, for each θ ∈ S , since m ( θ ) ∈ D ( θ ) and K ⊂ D + ( θ ), m ( θ ) cannot be in Int ( K ), hence m ( θ ) ∈ ∂K . θ We begin this section with some known results; we give the proofs for the sake of com-pleteness of the article. See Section 2, especially Figure 1, for the notation. We recallthat m is the midpoint of the chord bc and that v l | r = h ~m ′ l | r , ~u i . Proposition 4.1 . (a)
The curve m is the envelope of the family { ∆( α, θ ) , θ ∈ S } , i.e.,the vector products ~m ′ l ∧ ~u and ~m ′ r ∧ ~u vanish identically. (b) If b and c are regular points of ∂K then ~m ′ = v~u and v = h (cotan β + cotan γ ) . (10) In the case where b or c (or both) is a corner of ∂K , we have ~m ′ l = v l ~u , ~m ′ r = v r ~u , v l = h (cotan β l + cotan γ l ) , and v r = h (cotan β r + cotan γ r ) . (c) If b and c are regular then v is the signed radius of curvature of the curve m . If b or c (or both) is a corner then v l , resp. v r , is the signed radius of curvature on the left, resp.on the right, of m . Statement (a) is attributed to M. M. Day [7] by S. Tabachnikov. Formula (10) appearsin a similar form in [13].
Proof . We fix α ∈ ]0 ,
1[ and θ ∈ S ; we will not always indicate the dependence in α and θ of the functions b, c, m, ~u , etc. Let ε >
0, set M ( ε ) = ∆( α, θ ) ∩ ∆( α, θ + ε ) (see Figure 4),and consider the curvilinear triangles T b ( ε ) = ∆( α, θ ) − ∩ ∆( α, θ + ε ) + ∩ K , T c ( ε ) = ∆( α, θ ) + ∩ ∆( α, θ + ε ) − ∩ K. Proof of Proposition 4.1We have | ∆( α, θ ) − ∩ K | = | ∆( α, θ + ε ) − ∩ K | = α , hence0 = | T b ( ε ) | − | T c ( ε ) | = ε (cid:0) k b − M ( ε ) k − k c − M ( ε ) k (cid:1) + o ( ε ) . As a consequence, we obtain lim ε → + M ( ε ) = m . The case ε < ~b ′ r = h cotan β r ~u − h~u ′ . (11)For ε >
0, let B ( ε ) denote the intersection of the right-tangent to ∂K at b with the line∆( α, θ + ε ), see Figure 4. By definition of right-tangent, we have B ( ε ) = b ( θ + ε ) + o ( ε ).Set H ( ε ) = k b − M ( ε ) k , H ( ε ) = h B ( ε ) − b, ~u i , H ( ε ) = h M ( ε ) − B ( ε ) , ~u i , and H ( ε ) = h b − B ( ε ) , ~u ′ i . We obtain the following linear system in H , H , H H ( ε ) = H ( ε ) + H ( ε ) , H ( ε ) = H ( ε ) tan β r = H ( ε ) tan ε, giving B ( ε ) = b − tan β r tan ε tan β r + tan ε H ( ε ) ~u + tan ε tan β r + tan ε H ( ε ) ~u ′ . As a consequence, we obtain B ( ε ) = b − εH ( ε ) ~u + ε cotan β r H ( ε ) ~u ′ + o ( ε ). Since H ( ε ) = h + o (1), it follows that b ( θ + ε ) = B ( ε )+ o ( ε ) = b − εh~u + ε cotan β r h~u ′ + o ( ε ), yielding (11).Similarly, we have ~b ′ l = h cotan β l ~u − h~u ′ , ~c ′ l = h cotan γ l ~u + h~u ′ , and ~c ′ r = h cotan γ r ~u + h~u ′ .Statements (a) and (b) now follow from m = ( b + c ). Since the expression of v givenby (10) is a ruled function, statement (c) follows directly from Proposition 3.2 (a). Corollary 4.2 . For all θ ∈ S and all α ∈ ]0 , , we have lim θ → θ − v l ( θ ) = lim θ → θ − v r ( θ ) = v l ( θ ) and lim θ → θ +0 v l ( θ ) = lim θ → θ +0 v r ( θ ) = v r ( θ ) . (12) In other words, in the sense of the Pompeiu-Hausdorff distance, we have lim θ → θ − V ( θ ) = { v l ( θ ) } and lim θ → θ +0 V ( θ ) = { v r ( θ ) } . (13) Therefore, the function V is upper semi-continuous (for the inclusion) with respect to θ .Proof . Immediate, using (3), (4), (10), and the continuity of the cotangent function.The next statement will be used both in Sections 5 and 6. orollary 4.3 . For θ , θ ∈ S , with θ ≤ θ ≤ θ + π , we have conv (cid:0) V ( α, θ ) ∪ V ( α, θ ) (cid:1) ⊆ [ θ ≤ θ ≤ θ V ( α, θ ) . (14) Proof . Let v be in the above convex hull. If v ∈ V ( α, θ ) ∪ V ( α, θ ), there is nothing toprove. Otherwise assume, without loss of generality, that max V ( α, θ ) < min V ( α, θ )and set θ = sup { θ ≥ θ ; max V ( α, θ ) < v } . We have θ ≤ θ < θ ; for all θ ∈ ] θ , θ [,max V ( α, θ ) ≥ v ; and there exists a convergent sequence { θ n } n ∈ N tending to θ withmax V ( α, θ n ) < v . By (12), we obtain max V ( α, θ ) = v , hence v ∈ V ( α, θ ). In this section we fix α ∈ ]0 ,
1[ . Recall that θ is called regular if ∂K is C at b ( α, θ ) and c ( α, θ ), and singular otherwise. Definition 5.1 . (a) Let F ( α ) be the set of all θ ∈ S such that either v ( α, θ ) > θ is regular), or V ( α, θ ) ∩ ]0 , + ∞ [ = ∅ (if θ is singular); we call it the forwards set .(b) Similarly, let B ( α ) be the set of all θ ∈ S such that either v ( α, θ ) < θ is regular),or V ( α, θ ) ∩ ] − ∞ , = ∅ (if θ is singular); we call it the backwards set .(c) Finally, let Z ( α ) denote the set of all θ ∈ S such that either v ( α, θ ) = 0 (if θ isregular), or V ( α, θ ) contains 0 (if θ is singular); we call it the zero set .By the symmetry (7), we have (with the notation [ θ , θ ] = { θ ∈ S ; θ ≤ θ ≤ θ } for θ , θ ∈ S ) θ ∈ Z ( α ) ⇔ θ + π ∈ Z (1 − α ) and θ ∈ B ( α ) ⇔ θ + π ∈ F (1 − α ) . (15)By Proposition 4.1 (b), we have θ ∈ Z ( α ) ⇔ π ∈ [ β l + γ l , β r + γ r ] (or π ∈ [ β r + γ r , β l + γ l ]) . Since β l ≥ β r and γ r ≤ γ l , by (2), we deduce that, if θ ∈ Z ( α ) then b = b ( α, θ ) and c = c ( α, θ ) admit parallel supporting lines of K . If one of the points b, c or both isregular, the converse is also true; however, it can occur that β r + γ l ≤ π ≤ β l + γ r but π does not belong to [ β l + γ l , β r + γ r ] (or to [ β r + γ r , β l + γ l ]), and then V = [ v l , v r ] (or[ v r , v l ]) does not contain 0.In the case where ∂K is C , by Proposition 4.1, θ belongs to B ( α ) if and only if β ( α, θ ) + γ ( α, θ ) > π , i.e. , there exists a triangle T = abc , with one edge equal to thechord bc (with b = b ( α, θ ) , c = c ( α, θ )), the other two edges, ab and ac , not crossing theinterior of K , and which contains an amount 1 − α of K : | T ∩ K | = (1 − α ) | K | . In thecase where ∂K is not C , this latter condition is necessary but not always sufficient tohave θ ∈ B ( α ). However, we will see in Section 6 that this condition implies θ ∈ B ( α ′ )for all α ′ > α .Let us observe Z ( α ) and B ( α ), denoted Z and B here. These sets can be very compli-cated, even if ∂K is C . In Section 8 we present a construction which, to any prescribedclosed subset C of S , associates a C convex body K such that Z (cid:0) (cid:1) coincides with C ,up to countably many isolated points. We give next a brief description in the simplestcases.If θ is an isolated point of Z which does not belong to the closure of B , then m ( θ )is a point of zero curvature of m , but m is still C at θ . f θ < θ < θ + π and [ θ , θ ] ⊂ Z is an isolated connected component of Z , then m has a corner; i.e., it is not C but has two half-tangents: a left-tangent oriented either by ~u ( θ ) if ] θ − δ [ ⊂ F for small δ >
0, or by ~u ( θ + π ) if ] θ − δ [ ⊂ B , and a right-tangentoriented either by ~u ( θ ) or by ~u ( θ + π ). Moreover, in this situation, m ( α, θ ) is a localcenter of symmetry of ∂K : the arc of ∂K in the sector ∆( α, θ ) + ∩ ∆( α, θ ) − is symmetricto that in ∆( α, θ ) − ∩ ∆( α, θ ) + .If θ is an isolated point of Z and is the endpoint of both a segment ] θ , θ [ of F anda segment ] θ , θ [ of B , then m has a cusp at m ( θ ). Proposition 5.2 . Let α ∈ ]0 , . (a) The set Z ( α ) is closed in S . (b) If ∂K is C then the sets F ( α ) and B ( α ) are open in S and F ( α ) , B ( α ) , and Z ( α ) form a partition of S . (c) In the general case, the three sets F ( α ) ∩ B ( α ) , ∂F ( α ) , and ∂B ( α ) are subsets of Z ( α ) .Remark. If ∂K is not C then, in general, there exists α ∈ ]0 ,
1[ such that B ( α ) and F ( α )are not open. Actually, if c is a corner of ∂K , with half-tangents of directions θ , θ , and b ∈ ∂K is such that all supporting lines to K at b have directions in ] θ + π, θ + π [ , thenthe α -section (for some α ) passing through b and c has a direction θ ∈ F ( α ) ∩ B ( α ), butsuch that any θ < θ , θ close enough to θ , does not belong to B ( α ) and any θ > θ , θ close enough to θ , does not belong to F ( α ). Proof . (b) If ∂K is C then the functions β and γ are continuous with respect to θ ,hence also is v , so Z = v − ( { } ) is a closed subset of S , and F = v − ( ] − ∞ ,
0[ ) and B = v − ( ]0 , + ∞ [ ) are open. Furthermore they do not intersect and their union is S .(a) Let { θ n } n ∈ N be a sequence converging to θ , with θ n ∈ Z ( α ), i.e., ∈ V ( α, θ n ).Then a subsequence { θ n k } k ∈ N exists such that θ n k tends either to θ − or to θ +0 , say to θ − .By (13) we have lim k → + ∞ V ( α, θ n k ) = (cid:8) v l ( α, θ ) (cid:9) , hence v l ( α, θ ) = 0, and θ ∈ Z ( α ).(c) Let θ ∈ ∂F ( α ); then there exist sequences { θ n } n ∈ N in F ( α ) and { θ ′ n } n ∈ N in S \ F ( α ),both converging to θ . This means that V ( θ n ) ∩ ]0 , + ∞ [ = ∅ and V ( θ ′ n ) ⊂ ] −∞ ,
0] for every n ∈ N . By (14), there exists θ ′′ n ∈ [ θ n , θ ′ n ] (or θ ′′ n ∈ [ θ ′ n , θ n ]) such that 0 ∈ V ( θ ′′ n ) for each n , i.e., θ ′′ n ∈ Z ( α ). Since the sequence { θ ′′ n } n ∈ N tends to θ and Z ( α ) is closed, we obtain θ ∈ Z ( α ). The proof for ∂B ( α ) ⊆ Z ( α ) is similar. The proof for F ( α ) ∩ B ( α ) ⊆ Z ( α ) isobvious. α The functions b and c are left- and right-differentiable with respect to α ; one finds, e.g. , −→ (cid:0) ∂b∂α (cid:1) l = h ( ~u ′ + cotan β r ~u ). However, we will not use their differentiability in α , but onlytheir monotonicity. Since the function cotan is decreasing on ]0 , π [ , by (2) we immediatelyobtain the following statement, whose proof is omitted. Proposition 6.1 . The functions v, v l and v r are nonincreasing in α . More precisely, wehave max V ( α ′ , θ ) ≤ min V ( α, θ ) for all θ ∈ S and all < α < α ′ < . (16) ecall the sets F ( α ), B ( α ), and Z ( α ) introduced in Definition 5.1. Let I F = { α ∈ ]0 ,
1[ ; F ( α ) = ∅} ,I B = { α ∈ ]0 ,
1[ ; B ( α ) = ∅} ,I Z = { α ∈ ]0 ,
1[ ; Z ( α ) = ∅} . The symmetry (15) implies α ∈ I F ⇔ − α ∈ I B and α ∈ I Z ⇔ − α ∈ I Z . (17)Set α B = inf I B and α Z = inf I Z . (We do not consider inf I F , which is always equal to 0,see below). Theorem 6.2 . (a)
We have I B = ] α B , (and hence I F = ]0 , − α B [ by (17)). If ∂K is C , then I Z = [ α Z , − α Z ] ; otherwise I Z is one of the intervals [ α Z , − α Z ] or ] α Z , − α Z [ . (b) We have α Z ≤ α B ≤ . Moreover, if α Z = α B then ∂K contains two parallelsegments. (c) We have the following equivalences: i. α B = if and only if K is symmetric. ii. α Z = 0 if and only if ∂K contains a segment whose endpoints admit two parallelsupporting lines to K . iii. α B = 0 if and only if ∂K contains a segment whose endpoints admit two parallelsupporting lines to K , one of which intersecting ∂K at only one point.Remarks.
1. Notice the change of behaviour of Z and B : In Proposition 5.2, Z ( α ) isalways closed, whereas B ( α ) may be not open if ∂K is not C ; here I B is always open,whereas I Z may be not closed if ∂K is not C .2. An example of a planar convex body such that I Z is open is the quadrilateral OICJ in Figure 9, Section 8.9, for which one finds I Z = (cid:3) c , − c (cid:2) . Proof . (a) By Proposition 6.1, if 0 < α < α ′ < B ( α ) ⊆ B ( α ′ ). It follows that I B = ] α B ,
1[ or I B = [ α B , α B / ∈ I B .If α ∈ I B , then there exists θ ∈ S such that, say, v l ( α, θ ) < v r ( α, θ ) < α ′ < α and θ ′ < θ be such that b ( α ′ , θ ′ ) = b ( α, θ ); then by (12) , both v l ( α ′ , θ ′ ) and v r ( α ′ , θ ′ ) tend to v l ( α, θ ) as θ ′ → θ and α ′ → α , hence are negative for ( α ′ , θ ′ )close enough to ( α, θ ). This shows that α ′ ∈ I B . As a consequence, I B = ] α B ,
1[ .We now prove that I Z is convex. Let α < α ∈ I Z ; i.e. , there exist θ and θ in S such that 0 ∈ V ( α , θ ) ∩ V ( α , θ ). We may assume, without loss of generality,that θ ≤ θ ≤ θ + π . Let α ∈ ] α , α [. By (16), we have V ( α, θ ) ∩ R − = ∅ and V ( α, θ ) ∩ R + = ∅ hence, by (14), we have 0 ∈ V ( α, θ ) for some θ ∈ [ θ , θ ] (in the case θ = θ + π , both intervals [ θ , θ ] and [ θ , θ + 2 π ] suit). This shows that α ∈ I Z .(b) If α belongs neither to I F nor to I B , then we necessarily have V ( α, θ ) = { } for all θ ∈ S , hence the function m ( α, · ) is constant. This implies α = and K symmetric. Asa consequence, we have α B ≤ .We prove ∈ I Z , yielding I Z = ∅ . Let θ ∈ S arbitrary. By (7), 0 ∈ conv (cid:0) V (cid:0) , θ (cid:1) ∪ V (cid:0) , θ + π (cid:1)(cid:1) , hence by (14) 0 ∈ V (cid:0) , θ (cid:1) for some θ ∈ [ θ , θ + π ], hence Z (cid:0) (cid:1) = ∅ .We now prove α Z ≤ α B . If α B = , we are done; otherwise, let α ∈ I B , α ≤ . Then v l ( α, θ ) or v r ( α, θ ) is negative for some θ ∈ S , say v l ( α, θ ) <
0. From (6) and (16) it In fact (12) is stated for α fixed, but its proof uses only (3) and (10) — which can easily be adapted toour situation — and the continuity of the function cotan. ollows that v l ( α, θ + π ) = − v l (1 − α, θ ) ≥ − v l ( α, θ ) >
0, hence by continuity there exists θ ∈ ] θ, θ + π [ such that v l ( α, θ ) = 0, so θ ∈ Z ( α ), and α ∈ I Z .If moreover α Z < α B , then consider α Z < α < α ′ < α B ≤ , hence α, α ′ ∈ I Z \ I B .Let θ ∈ Z ( α ), b = b ( α, θ ), c = c ( α, θ ), and let D b , D c be two parallel supporting lines to K at b and c respectively, see Figure 5. Consider now b ′ = b ( α ′ , θ ) (for the same θ ) and Figure 5:
Proof of Theorem 6.2 (b) c ′ = c ( α ′ , θ ). Since α < α ′ , ∆( α ′ , θ ) is in the interior of ∆ + ( α, θ ). Since B ( α ′ ) = ∅ , theredo not exist supporting lines to K at b ′ and c ′ crossing in ∆ + ( α ′ , θ ). As a consequence, b ′ must lie on D b and c ′ on D c , and the segments [ b ′ , b ] and [ c ′ , c ] are on ∂K .(c) i. If α B = then by (a) / ∈ I F ∪ I B hence, as already said in the proof of (b), K issymmetric. The converse is obvious.(c) ii. and iii. If α Z = 0 then there exists a sequence { α n } n ∈ N tending to 0, with α n ∈ I Z ; i.e. , such that for any n there exists θ n ∈ S with 0 ∈ V ( α n , θ n ). By compactness, we mayassume without loss of generality that the sequence { θ n } n ∈ N converges to some θ ∈ S .Let b n = b ( α n , θ n ) and c n = c ( α n , θ n ). By continuity, the sequences { b n } n ∈ N and { c n } n ∈ N converge to some b , resp. c ∈ ∂K . Since the width of K satisfies 0 < w ( K ) ≤ k b n − c n k ,we have b = c . Since 0 ∈ V ( α n , θ n ), there are parallel supporting lines to K at b n and c n for each n . Without loss of generality, we also assume that these lines converge,yielding two parallel supporting lines at b and c , denoted D b and D c . Since the sequenceof lines (cid:8) ∆( α n , θ n ) (cid:9) n ∈ N tends to the oriented line through b and c , denoted by ( bc ), and | ∆ − ( α n , θ n ) ∩ K | = α n | K | →
0, we obtain | ( bc ) − ∩ K | = 0, hence the segment [ b, c ] is on ∂K .Conversely, if [ b, c ] ⊂ ∂K and b = c admit two parallel supporting lines to K , denotedby D b and D c , let θ denote the direction of ( bc ); i.e., θ is such that ~bc = k~u ( θ ) forsome k >
0. The angles β n = β (cid:0) n , θ (cid:1) and γ n = γ (cid:0) n , θ (cid:1) satisfy β n + γ n ≥ π , hence θ ∈ B (cid:0) n (cid:1) ∪ Z (cid:0) n (cid:1) , so B (cid:0) n (cid:1) ∪ Z (cid:0) n (cid:1) = ∅ , n ∈ I B ∪ I Z , and 0 ≤ min( α Z , α B ) = 0, hence α Z = 0 since α Z ≤ α B .If one of the supporting lines above, say D b , intersects ∂K only at b , then b n / ∈ D b ,yielding β n + γ n > π , hence n ∈ I B , showing α B = 0. Conversely, if α B = 0, then theformer points b n and c n admit supporting lines which cross in ∆ + ( α n , θ n ) (see commentafter Definition 5.1). By contradiction, if both D b ∩ ∂K and D c ∩ ∂K contained morethan b , resp. c , then for n large enough we would have b n ∈ D b and c n ∈ D c . Then for m, n such that b m ∈ ] b n , b [ and c m ∈ ] c n , c [ , the only supporting lines at b m , c m could be D b and D c , which do not cross, a contradiction. The α -core In this section we compare the boundary ∂K α with the image of m ( α, · ), denoted by m ∗ α ,for α ∈ ]0 ,
1[ . The function of θ , m ( α, · ), is a tour in the sense of Definition 3.1, with R l = v l ( α, · ) and R r = v r ( α, · ), and core K α . In the case α = , m (cid:0) , · (cid:1) is also a doublehalf-tour.Then, by Proposition 3.2 (b) and (c) and Theorem 6.2 (a), m ∗ α is of class C if 0 <α < α Z or 1 − α Z < α <
1, and is not C if α B < α < − α B . In the case α = α Z , m ∗ α may or may not be C . However, if α Z = α B , we will see that m ∗ α is not C also for α Z < α ≤ α B .By Proposition 3.2 (d), K α is strictly convex (or one point or empty), and by Propo-sition 3.2 (e) we have m ∗ α = ∂K α if and only if B ( α ) is empty. By Theorem 6.2 (a), wethen have m ∗ α = ∂K α ⇔ α ≤ α B . (18)Besides, the function α K α is continuous (for the Pompeiu-Hausdorff distance oncompact sets in the plane) and decreasing (with respect to inclusion): if α < α ′ , then K α ′ ( K α . Since K α = ∅ if α > , there exists a value α K ≤ such that • K α is strictly convex with a nonempty interior if 0 < α < α K , • K α K is a single point denoted by T , and • K α is empty if α K < α < α K is at least α B . It is noticeable that, when K α K is a single point,this point is not necessarily the mass center of K , see Section 8.9.We end this section with the following statement, which gathers the last results. Proposition 7.1 . (a) If α Z < α < − α Z , then m ∗ α is not C . (b) We have ≤ α K ≤ , with first equality if and only if K is a triangle, and secondequality if and only if K is symmetric. (c) If K is non-symmetric, then α B < α K (whereas for K symmetric we have α B = α K = ).Proof . (a) It remains to prove that, if α Z < α ≤ α B , then m ∗ α is not C . Assume α Z < α B ;by Theorem 6.2 (b) and its proof, there exist two parallel segments on ∂K , denoted by[ a, b ] and [ d, c ] with abcd in convex position, such that the line oriented by ~bc , denoted by D , is an α Z -section of K . Let D denote the line oriented by ~ad . Then D is a β -sectionfor some β ≥ α B . For any α ∈ ] α Z , α B ], α > α Z but close to it, there is an α -section,denoted by ∆( α, θ ), passing through c and crossing [ a, b ] and an α -section, denoted by∆( α, θ ), passing through b and crossing [ d, c ], see Figure 6. These α -sections intersectat some point P ∈ (cid:2) ( a + d ) , ( b + c ) (cid:2) . Then we have m ( α, θ ) = P for all θ ∈ [ θ , θ ], ~m ′ l ( α, θ ) collinear to ~P c = k ~P c k ~u θ , and ~m ′ r ( α, θ ) collinear to ~bP = k ~bP k ~u θ . Since θ = θ = θ + π , m ( α, · ) is not C at P . If α is not close to α Z , ∆( α, θ ) (passing through c ) will not cross [ a, b ], and ∆( α, θ ) (passing through b ) will not cross [ d, c ]. Nevertheless,the argument given above holds as well, except that ∆( α, θ ) and ∆( α, θ ) will not crosson (cid:2) ( a + d ) , ( b + c (cid:2) .(b) Suppose α K = . Let T be the unique point of K / . With the notation for theproof of Proposition 4.1 (a), see Figure 4, since every half-section contains T , we have M ( θ, θ ′ ) = T for all θ = θ ′ . Now this proof implies that (cid:12)(cid:12) θ − θ ′ (cid:12)(cid:12) ( k b ( θ ) − T k − k c ( θ ) − T k ) = O (cid:0) (cid:12)(cid:12) θ − θ ′ (cid:12)(cid:12) (cid:1) , Proof of Proposition 7.1 (a)for all θ < θ ′ . Letting θ ′ tend to θ , this shows that b ( θ ) and c ( θ ) are symmetric about T .The fact that α K ≥ and that α K = if and only if K is a triangle is well known;see, e.g. , [23]. It reduces to the following statement which we prove below. The mass center G of K is in ∆ + (cid:0) , θ (cid:1) for all θ . Moreover, there exists θ ∈ S suchthat ∆( , θ ) contains G if and only if K is a triangle. Fix θ , let ∆ = ∆ (cid:0) , θ (cid:1) , and consider the frame (cid:0) b = b ( θ ) , ~u = ~u ( θ ) , ~v = ~u ′ ( θ ) (cid:1) . Wehave to prove that y -coordinate G y of G is nonnegative and that this coordinate vanishesif and only if K is a triangle. Let D b and D c be two supporting lines of K at the points b and c = c ( θ ), see Figure 7. These lines together with the segment [ b, c ] define two convex Figure 7:
Proof of Proposition 7.1 (b). The triangle V = L ∪ U is in bold.sets C + = conv (cid:0) (∆ + ∩ D b ) ∪ (∆ + ∩ D c ) (cid:1) and C − = conv (cid:0) (∆ − ∩ D b ) ∪ (∆ − ∩ D c ) (cid:1) , one oneach side of ∆. The set K + = K ∩ ∆ + is included in C + and the set K − = K ∩ ∆ − isincluded in C − . Let a ′ ∈ D b ∩ ∆ − and a ′′ ∈ D b ∩ ∆ − be such that the triangles conv( a ′ , b, c )and conv( a ′′ , b, c ) have an area equal to | K | . Then the line ( a ′ a ′′ ) is parallel to ∆ andfor every a ∈ [ a ′ , a ′′ ] the triangle conv( a, b, c ) also has an area equal to | K | . For anysuch a , let b ′ ∈ ∂K ∩ [ a, b [ and c ′ ∈ ∂K ∩ [ a, c [. By continuity, one can choose a such thatthe oriented line ∆ ′ = ( b ′ c ′ ) is parallel to ∆. For this a , let L = conv( a, b, c ). We have K − \ L ⊂ ∆ ′ + and L \ K − ⊂ ∆ ′− , therefore the y -coordinate of the mass center G − of K − is larger than or equal to the y -coordinate of the mass center of L , the equality holdingonly when K − is a triangle. Furthermore, since a ∈ C − , the convex set C + is included in he convex set C ′ + = conv (cid:0) (∆ + ∩ ( ab ) ∪ (∆ + ∩ ( ac ) (cid:1) .Let U be the trapezoid of area | K | defined by the lines ( ab ), ( ac ), ∆ and a line∆ ′′ ⊂ ∆ + parallel to ∆. Since K + \ U ⊂ ∆ ′′ + and U \ K + ⊂ ∆ ′′− , the y -coordinate ofthe mass center G + of K + is larger than or equal to the y -coordinate of the mass centerof U , the equality holding only when K + = U . It follows that G y is larger than or equalto the y -coordinate of the mass center of the union V of L and U , with equality only if K = V . Since the mass center of V is on ∆ we are done.(c) Suppose that α B = α K . By (18), we obtain that m ( α K , · ) is constant and equal to T . It follows that, for all θ ∈ S , T is the middle of a chord of direction θ , hence K issymmetric about T . It is easy to check that, if an α -section crosses two non-parallel segments of ∂K ,then the corresponding middle of chord m ( α ) lies on an arc of hyperbola asymptotic tothe lines extending these segments. In particular, if K is a convex polygon, then for all α ∈ ]0 ,
1[ the curve m ( α, · ) is entirely made of arcs of hyperbolae. Of course, not onlysegments yield arcs of hyperbolae. One can check for instance that two arcs of hyperbolaeon ∂K also give an arc of hyperbola for m ( α, · ), if α is small enough. We saw that an envelope of α -sections is a tour in the sense of Definition 3.1.Conversely, which tours are envelopes of some α -sections m ( α, K ), for which convex bodies K and for which 0 < α <
1? For example, the astroid on the left of Figure 2 given by θ ( − cos θ, sin θ ), i.e., with R ( θ ) = sin(2 θ ), cannot be such an envelop. Moregenerally, following a remark by D. Fuchs and S. Tabachnikov in [11], a tour with acommon tangent at two different points cannot be such an envelop, since each pointwould have to be the middle of the chord. The KAM theory applied to dual billiards shows that, if m is a tour of class C (hence strictly convex), then there exist convex bodies K with ∂K arbitrarily close to m such that m = m ( α, K ) (hence for some α arbitrarily close to 0) and there exist convexbodies K with ∂K arbitrarily close to infinity such that m = m ( α, K ) (hence for some α arbitrarily close to ). The curves ∂K are invariant torii of the dual billiard. According toE. Gutkin et A. Katok [13], the works of J. Moser [20] and R. Douady [7] prove that thesecurves are convex. Notice the following apparent paradox. By Theorem 6.2(c)i., if K isnon-symmetric then, for α close to , m ( α ) is non-convex. This apparently contradictsthe above results which imply that, given a strictly convex C non-symmetric curve, thereexist values of α arbitrarily close to and convex bodies K , necessarily non-symmetric,such that m = m ( α, K ). However, these curves have envelops m ( β, K ) with cusps forsome other β ∈ ] α, [, although α is arbitrarily close to . There is no real contradiction.Several questions remain open: Among tours which present cusps, which ones areenvelopes of sections? For instance, does it exist a symmetric curve with cusps which isthe α -envelope of some (necessarily non-symmetric) convex body? Also here, does it exista non-symmetric convex body K having a symmetric convex envelope m ( K, α, · ) for some α ∈ ]0 ,
1[ ?
The link between envelopes of α -sections and dual billiards yields a simple proof forthe following (also simple) fact:The only convex bodies which have a circular envelope (or an elliptic one, since this isthe same question modulo an affine transformation) are the discs having the same centers s the envelop. Actually, if the billiard table is a circle, then the billiard map is integrable . This means that each orbit remains on a circle centered at the origin, and is eitherperiodic or dense in this circle, depending whether the angle between the two tangentsfrom the starting point at the table is in π Q or not. If the convex body K were not a disc,then its boundary would cross at least one circle with dense orbits, hence would containat least one dense orbit on the latter circle, hence the whole circle, a contradiction. Let C be a closed subset of S . We construct below a C convex curve which is theboundary of a convex body for which Z (cid:0) (cid:1) = C ∪ S , for some countable set S .As convex curve, we start with the unit circle and will deform it. Since C is closed, S \ C is a countable union of open intervals. Let ] θ , θ [ be one of them. Between θ and θ , we deform the circle into a convex C curve arbitrarily close to the union of twosegments, one [ e iθ , p ] tangent to the circle at e iθ , the other [ e iθ , p ], such that the arearemains unchanged. (For convenience, here we use the notation of complex numbers.)We do the same symmetrically with respect to the line passing through the origin andof direction ( θ + θ + π ); i.e. , we choose a C curve close to [ e i ( θ + π ) , p ′ ] tangent to thecircle at e i ( θ + π ) , and to [ e i ( θ + π ) , p ′ ], with p ′ = p e i ( θ + θ + π ) , see Figure 8.Observe that the curve is no longer (centrally-)symmetric. If the curve is chosen withlarge curvature near the points p and p ′ , then the interval ] θ , θ [ contains only two valuesof θ for which the half-section cuts the curve in a chord with two parallel tangents; thesechords have one endpoint near p , resp. near p ′ , and do not contain the origin anymore.Doing this in any connected component of S \ C , we obtain a convex C curve suchthat any θ ∈ C has a halving chord with parallel tangents and all but two values of θ ineach connected component of S \ C have a halving chord without parallel tangents. Figure 8:
A construction of C convex body with (almost) prescribed zero set It is easy to construct K L such that L α ⊂ K α for some value of α ; for example, L is an equilateral triangle, K the inscribed circle of L , and α ∈ (cid:2) , (cid:3) .Our calculation shows that this holds for all α ∈ (cid:2) α , (cid:3) , where α ≈ . − (1 − √ − α ) = cos t and α = π ( t − cos t sin t ). Actually, if the radius of K is R (hence the height of L is 3 R ), K α is a disc of radius R α = R cos t such that α = π ( t − cos t sin t ). Then ∂L α is made of three arcs of hyperbolae. In a (non-orthonormal) framewhere the triangle has vertices (0 , , , xy = α and x (1 − x − y ) = α , hence they cross at a point of abscissa x c = (1 − √ − α )) A classical conjecture states that only ellipses have a dual billiard map which is integrable. his proves that the three arcs of ∂L α cross at a distance R − R (1 − √ − α ) from thecenter.Is it true that, for every value of α ∈ ]0 , [ there exists K ( L such that L α ⊂ K α ? (Itcannot exist K, L , K ( L , independent of α such that L α ⊂ K α for all small α > α > K, L , K ( L , and all α < α , L α K α ? Is α ≈ . I.e., is it optimal when K is a disc and L a triangle? Or, instead, is it optimal when K is an affinely regularhexagon inscribed in L , a triangle? The following questions have been asked by Jin-ichi Itoh, whom we would like tothank for his interest in our work. Let w ( K ) denote the width of a convex body K , ⊘ ( K )its diameter, r ( K ) its inradius, and R ( K ) its circumradius.Do we have w ( K ) ⊘ ( K ) ≤ w ( K α ) ⊘ ( K α ) for all α < ?Do we have r ( K ) R ( K ) ≤ r ( K α ) R ( K α ) for all α < ?Both answers are “no”, as shown by the following counter-examples.For the first question, if K is the unit Reuleaux triangle, ⊘ ( K ) = w ( K ) = 1, and α > ⊘ ( K α ) = 1 − O (cid:0) α / (cid:1) , whereas w ( K α ) < − α / < ⊘ ( K α ).For the second question, consider two small arcs of the circle of center 0 and radius 2near the x -axis, two small arcs of the circle of center 0 and radius 1 near the y -axis, andchoose for K the convex hull of the union of these four arcs.In this manner we have r ( K ) R ( K ) = . However, we have r ( K α ) = cos t , where t is suchthat α = | K | ( t − sin t cos t ) = | K | t + O ( t ) , and R ( K α ) = 2 cos t , where t is suchthat α = | K | ( t − sin t cos t ) = | K | t + O ( t ) , hence t < t , so r ( K α ) R ( K α ) < .Whether there is a constant k > w ( K ) ⊘ ( K ) ≤ k w ( K α ) ⊘ ( K α ) for all convex bodies K and all α , idem for r ( K ) R ( K ) , and which constant is optimal, seems to be another interestingquestion. There is a tight link between the region where appear cusps, as described in [11],and the regions M and M described by T. Zamfirescu in [31]. Actually, let K be aplanar convex body and, for each θ ∈ S , consider the so-called midcurve , i.e. , the locusof midpoints of all chords of direction θ . If ∂K is C and strictly convex then the familyof all these midcurves for all θ ∈ S is a continuous family in the sense of Gr¨unbaum [12],for which general results of [31] are at our disposal.In our situation, the regions M and M of [31] are the loci of points of Int ( K ) which aremiddles of at least 2, resp. 3, different chords. If K is symmetric then M = M = { g ( K ) } ,the mass center of K . If K is not symmetric then it seems that we haveInt ( M ) = M = Int (cid:16) [ α ∈ ]0 , [ ( m ∗ α \ ∂K α ) (cid:17) . (19)Already (19) can be verified when K is a quadrilateral. For instance, the polygonal curveof all cusps described in Figure 11.15 of [11] is indeed the boundary of M .Besides, when K is a quadrilateral, a detailed analysis shows that the unique point T of K α K is in M and that the three chords bisected by T are all α K -sections. Thisproperty of T is likely to be true in the general case. Conversely, is T the only pointof M bisecting three α -sections with the same α ? This has been checked in the case ofquadrilaterals. .9 We present here an explicit example showing that the unique point T of K α K is notnecessarily the mass center G of K . The strategy of proof is to use a quadrilateral and toshow that the three chords bisected by G are α -sections for different values of α , showingthat T = G by Subsection 8.8.Given c >
1, let K = OIJ C denote the quadrilateral conv(
O, I, J, C ), with O =(0 , , I = (1 , , J = (0 , , C = ( c, c ). The area of K is | K | = k −→ IJ k k −→ OC k = c and its mass center is G = (cid:0) c +16 , c +16 (cid:1) (the midpoint of the segment determined bythe centers of mass of the triangles OJ C and
OIC ). The point G is the midpoint ofexactly three chords, AB , EF , and E ′ F ′ , with A, B on the line of equation x + y = c +13 , E ∈ OJ , F ∈ IC , and E ′ ∈ OI , F ′ ∈ J C , see Figure 9. These chords divide K in ✲✻ (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0)(cid:0)(cid:0)✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✁✧✧✧✧✧✧✧✧✧✧✧✧✧✔✔✔✔✔✔✔✔✔✔✔✔✔❅❅❅❅❅❅❅ xy r O r I r J r C r G r H r E r F r E ′ r F ′ r B r A Figure 9:
The α -center is not the mass center; here c = 2proportion ( α AB , − α AB ), resp. ( α EF , − α EF ) and ( α E ′ F ′ , − α E ′ F ′ ) (by convention,0 < α XY ≤ ). By symmetry, we have α EF = α E ′ F ′ , but there is no reason that α EF = α AB ; as we will see, this is indeed not the case.The triangle ABC is the image of
IJ C by the dilation of center C sending the point H = (cid:0) , (cid:1) to G , hence of factor k −→ GC kk −→ HC k = c − c − . With | IJ C | = c − , this gives | ABC | = (4 c − c − , so α AB = (4 c − c (2 c − .Since F is on the line IC of equation c ( x −
1) = ( c − y , one finds F = (cid:0) c +13 , c (cid:1) ,hence E = (cid:0) , (cid:1) . The area of the triangle EIF is det( −→ EI −→ EF ) = (cid:12)(cid:12)(cid:12)(cid:12) c +13 −
13 2 c − (cid:12)(cid:12)(cid:12)(cid:12) = c − ,hence the area of the quadrilateral OIF E is | OIF E | = | OIE | + | EIF | = + c − = c +118 .This gives α EF = c +118 c = α AB , hence G = T .More generally, given a planar convex body G , we consider all chords the midpoint ofwhich coincides to the mass center G of K . It is known [30] that there are at least threedifferent such chords; these chords are α -sections for different values of α ranging fromsome α min ( K ) to some α max ( K ). Then the quotient α min ( K ) α max ( K ) measures in some sense theasymmetry of the body K . It is equal to 1 if K is (centrally) symmetric, or if K hasanother symmetry which ensures that G is the only affine-invariant point, e.g. for K aregular polygon, but it differs from 1 in general. One can see that the minimum of thisquotient is achieved for at least one affine class of convex bodies, and it would be worthy todetermine the shape of these bodies, and to compute the corresponding minimum. In thecase of all quadrilaterals, using Maple, we found that the minimum is attained preciselyfor a quadrilateral of our previous family, for c = and with α min ( K ) α max ( K ) = . The samequestions can be asked in higher dimensions, replacing the midpoint of a chord by themass center of an α -section. We end this section with our main conjecture. onjecture 8.1 . For any convex bodies
K, L with K ⊂ L , and any α ∈ ]0 , [ , thereexists an α -section of L which is a β -section of K for some β ≤ α . This conjecture has been recently proven in [10] in the case of planar convex bodies.Another natural related question, the answer of which turns out to be negative, is thefollowing.For K ⊂ L convex bodies, does there always exist a half-section ∆ of L such that | K | length(∆ ∩ L ) ≤ | L | length(∆ ∩ K ) ?For a counter-example, consider for K a thin pentagon of width varying from ε on eachside to 2 ε in the middle, placed at the basis of L , a triangle of heigth 1, see Figure 10. A ✡✡✡✡✡✡✡✡✡✡✡❏❏❏❏❏❏❏❏❏❏❏✡✡✡✡✡✡✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❏❏❏❏❏❏ Figure 10:
In bold, the body K ; in thin, the body L ; here ε = detailed analysis shows that, for every half-section ∆ of L , one has | K | length(∆ ∩ L ) > | L | length(∆ ∩ K ). Acknowledgements.
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95 (1967), 20–33.[33] E. Werner, Floating bodies and illumination bodies, Proceedings of the Conference“Integral Geormetry and Convexity”, Wuhan 2004, World Scientific, Singapore.Addresses of the authors:Nicolas Chevallier and Augustin FruchardLaboratoire de Math¨ı¿ matiques, Informatique et ApplicationsFacult¨ı¿ des Sciences et TechniquesUniversit¨ı¿ de Haute Alsace2 rue des Fr¨ı¿ res Lumi¨ı¿ re68093 Mulhouse cedex, FRANCEE-mails: [email protected], [email protected]
Costin VˆılcuSimion Stoilow Institute of Mathematics of the Romanian Academy,P.O. Box 1-764, Bucharest 70700, ROMANIAE-mail: