This section is devoted to the proofs of the main results. As announced in the Introduction, as a first step we address the problemAN,N ∈N∗, of area minimization restricted to the class of Reuleaux polygons with at most2N+ 1sides. We will prove the following.
Theorem4.1. — The area minimization problem restricted to the family of Reuleaux polygons with at most2N+ 1sides,N ∈N∗, has the following solution: ifN < N(r), then there is no admissible shape forAN(r); otherwise, ifN >N(r), the unique (up to rigid motion) minimizer ofAN(r)isΩ(r), whereN(r)andΩ(r)are the function and the shape introduced in Definition 3.10.
In order to prove Theorem 4.1, we need to compute the area of a rigid shape.
To this aim, we split a shape withM sides intoM subdomains, by connecting with straight segments the origin to the vertexes. As previously, the origin is put at the center of the minimal annulus. The elements of this partition can be regrouped as triples of subdomains associated to clusters and subdomains associated to extremal
arcs. Examples of triples of subdomains associated with clusters are the gray regions in Figure 6 : on the left, one triple; on the right, two triples. In the next lemma we provide a formula for the areas of these subdomains.
Lemma4.2. — Letrbe the inradius. Then the area of a triple of subdomains associated to a cluster of parameterhis
(4.1) F(r, h) := (1−r)2sinhcosh+a−sina 2
+ (1−r) cos(a/2)−(1−r) cosh
sin(h+`) +b−sinb, where`=`(r),a=a(r, h)andb=b(r, h)are the functions introduced in(3.2),(3.3), and (3.4), respectively.
Proof. — The areaF(r, h)is the sum of two terms:F(r, h) =|A|+ 2|B|, where Ais the subdomain with boundary arc of length aand B is one of the two subdomains with boundary arc of lengthb (which clearly have the same area). Each of them can be further decomposed as a triangle of the form OPjPj+2 and a portion of disk. For the subdomainA, the triangle is isosceles: the two sides which meet atOhave length 1−rand meet with an angle of2h, therefore the area is
(1−r)2sinhcosh.
The area of the remaining part can be computed by difference, as the area of the circular sector with vertexes PjPj+1Pj+2 and the triangle with the same vertexes.
The result is
a−sina
2 .
Let us now considerB. The triangle inB (not isosceles) has the following structure:
the two sides which meet atOhave length1−randcos(a/2)−(1−r) cosh, respectively, and form an angle of amplitude`+h; therefore its area is
1
2(1−r) cos(a/2)−(1−r) cosh
sin(h+`).
As already done forA, it is immediate to check that the remaining part inBhas area b−sinb
2 .
By summing up the contributions we find (4.1).
Remark4.3. — Notice that the formula above is valid also for h= 0 or `, with the appropriate interpretation. As already noticed in Remarks 3.4 and 3.9, whenh= 0, the cluster reduces to a single extremal arc. The formula above at 0 gives
F(r,0) = (1−r)2sin`cos`+`−sin`
2 ;
which is the area of the subdomain bounded by an extremal arc and the two segments joining its endpoints to the origin. Similarly, whenh=`we have three extremal arcs, which is in accordance to
F(r, `(r)) = 3F(r,0).
The properties ofF are summarized in the following.
Proposition4.4. — The first and second derivatives of F with respect to the second variable are given by wherea=a(r, h) is the function introduced in (3.3). In particular,
∂2F(r, h)
∂h2 <0
for any h∈[0, `(r)],`(r)being the function introduced in (3.2).
Proof. — Throughout the proofris fixed, therefore we omit the dependence on it. In particular,F,a, andb, introduced in (4.1), (3.3), (3.4), respectively, will be regarded as functions of the sole variableh, and their derivatives will be denoted simply by a prime.
We will use the following formulas, which can be deduced from tan(`/2) = p4(1−r)2−1(cf. (3.2)): From the definition of aandb, we have
(4.6) a0= 2(1−r) cosh
cos(a/2), b0 = 1−(1−r) cosh cos(a/2). Differentiating F and using (4.6) yields:
F0(h) = (1−r)2cos(2h) + (1−r)(1−cosa) cosh
Using (3.3), (3.4), and (4.4), we obtain
cosa= 1−2(1−r)2sin2h, cos(a/2) = 1−(1−r)2sin2h cos(a/2) ,
and These computations allow to simplify the expression above ofF0 and to get (4.2).
Differentiating one more time (4.2) we get
F00(h) =−8(1−r)2coshsinh+2(1−r)3sinhcos2h
If we prove thatAandB are negative, we are done.
Sinceh7→a(h)is increasing, both terms in A are increasing. ThereforeA(r, h)6 A(r, `). Using (4.4) and (4.5) we get
Now, comparing theirsin, it is immediate that, for anyh,a/26(1−r)h. Therefore, B1(r, h)6B2(r, h) =−35
9 (1−r) coshcos((1−r)h) + 1 +(1−r)2
2 (5 cos(2h) + 1).
We now compute the three first derivatives ofh7→B2(r, h). It comes, after lineariza-tion
Since r4+ (2−r)4 6 56/9 and 1−r > 1/2 we conclude that d3B2/dh3 > 0 and thendB2/dhis convex inh, moreover it vanishes at0. Thus,dB2/dhis either always positive or always negative or negative and then positive (and this is actually the case). In any case, we see that
B2(r, h)6max (B2(r,0), B2(r, `)). Now we see that
B2(r,0) = 3(1−r)2−35
9 (1−r) + 16− 7 36 <0.
It remains to estimateB2(r, `). For that purpose, we claim the following:
(4.7) cos((1−r)`)>11 5 −12
5 (1−r) = 12 5 r−1
5.
Recalling the relation (3.2) between ` and r, the validity of (4.7) is related to the positivity of the auxiliary function The second derivative ofψreads
ψ00(r) =− 144(12r−1)
(25−(12r−1)2)3/2 + 2
(1−r)(4(1−r)2−1)3/2 and is negative in [1−1/√
3,1/2]. In particularψ is concave and ψ(r)>min(ψ(1−1/√
This leads to consider the polynomial Q(x) =−19 This polynomial is negative in[1/2,1/√
3], implying thatB2is negative too.
Proof of Theorem4.1. — We begin by noticing that if N < N(r), then the class of admissible shapes is empty: assume by contradiction that there exists a Reuleaux polygon contained into the annulusB1−r(0)rBr(0)withM <2N(r) + 1sides. Each arc of the boundary has length at most`(r), therefore, imposing that the perimeter isπand recalling the definition (3.7) ofN(r), we get
π6M `(r)6(2N(r)−1)`(r) =
Let nowN >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 admissible Blaschke deformation, to decrease the area. Thus, it remains to minimize the area among rigid shapes. We have seen in Proposition 3.5 that these rigid shapes are composed of extremal arcs and clusters.
Step 2. — Let us write an area formula for a rigid shapeΩ. Connecting with straight segments the origin to the vertexes, we split Ω into subdomains, which can be re-grouped as triples of subdomains associated to clusters (see also Figure 6) and sub-domains associated to extremal arcs. According to the notation used in Remark 3.9, all the subdomains can be regarded associated to clusters, allowing the parametershi
to vary in the closed interval[0, `(r)],i= 1, . . . ,m, for a suitablee me ∈N. In view of Lemma 4.2 and Remark 4.3, we infer that the total area is
|Ω|=
me
X
i=1
F(r, hi).
Notice that the area does not explicitly depend on the relative position of the clusters (this dependence is enclosed into the relation among the lengths). The necessary con-dition (3.6) gives a restriction on the possible values ofm: since everye hi is between0 and`(r), we infer that2hi+`(r)∈[`(r),3`(r)], so that, summing overifrom1tom,e we get
(4.8) π
3`(r) 6me 6j π
`(r) k.
In particular, this implies that the number of sides of a rigid configuration cannot be arbitrarily large, but it is bounded by a quantity depending only onr. We infer that the sequence of minima {AN(r)}N>N(r) is constant after a finite number of values (depending onr). A priori, the first terms of the sequence could be different. In the next step we show that, actually, the sequence is constant inN.
Step 3. — Let us optimize the area when me is fixed. In view of the previous step, we are led to minimize the function
F(h1, . . . , h
me) :=
me
X
i=1
F(r, hi), over the set
C :=n
(h1, . . . , hme)∈[0, `(r)]me :Pme
i=1[2hi+`(r)] =πo .
In that way, we transform a geometric problem into an analytic one which might have solutions that do not correspond to real geometric shapes. It turns out, as we will see below, that the minimizer 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 of Proposition 4.4, the functionF is strictly concave, therefore it attains a minimum on extremal points ofC. The extremal points ofC lie on the edges of the hypercube,
namely (up to relabeling)h1∈[0, `]andh2, . . . , hme ∈ {0, `}. Without loss of general-ity, we may label thehis in such a way thath2, . . . , hq= 0and hq+1, . . . , hme =`.
We claim the following facts:
(i) forme fixed, the extremal point ofC is unique;
(ii) for a fixedr, the minimum ofF does not depend onme in the range (4.8).
The case in whichris the inradius of some regular Reuleaux polygon is trivial: in view of (3.6), the parameter h1 has to belong to{0, `(r)} and, again by (3.6), no matter how the sides are regrouped (one by one whenh= 0, three by three whenh=`(r)), they are necessarily2N(r) + 1, whereN(r)is the number introduced in (3.7).
In all the other cases,h1 lies necessary between0 and`(r), strictly. A first conse-quence is that the number of sides is3 + (q−1) + 3(me −q). Since it is odd, we infer
More precisely, taking into account thath1/`(r)∈(0,1), we get q=q(m) := 1 +e j3
2me − π 2`(r)
k. Using again (3.6), we infer thath1 is given by
h1=`(r)(1−δ),
All in all, once fixedm,e h1 andqare determined. This concludes the proof of (i).
Notice that if we replace me byme + 2 (as already noticedme has to be odd), the value ofδdoes not change. This allows us to writeh1, without the dependence onm.e Therefore, in order to prove (ii), it is enough to show that the number of sides of length`(r)does not depend onm:e
q(m)e −1 + 3(me −q(m)) = 3e me −2q(m)e −1 = 3me −2
Step 4. — In view of the previous step, we immediately get that the optimal shape associated to the inradiusrof a regular Reuleaux polygon, is the Reuleaux polygon itself, for everyN >N(r). Whenris 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 `(r). As already underlined in Defini-tion 3.10, these properties characterize the setΩ(r), and the proof of the theorem is concluded. Note that 2bπ/(2`(r))−3/2c+ 2 (i.e., the number of sides of length`(r)
found in Step 3) is equal toN(r)−2, implying that the total number of sides of the
optimal shape is2N(r) + 1, as expected.
We are now in a position to prove the main results, about the characterization of minimizers and the continuity of minima and minimizers with respect tor.
Proof of Theorem1.2. — In view of the density of the Reuleaux polygons in the class of constant width sets see [3] or [4], we infer that
A(r) = inf
N>1AN(r).
In view of Theorem 4.1, we infer that the sequenceAN(r)is finite and constant after N(r), so thatA(r) = infN>1AN(r) =|Ω(r)|. The other statements follow from the characterization ofΩ(r)(see Definition 3.10 and Proposition 3.8).
Proof of Proposition1.4. — In view of Theorem 1.2, its proof, and Definition 3.10, A(r)can be computed by dividing the optimal shapeΩ(r)into2N(r)+1subdomains, obtained by joining with segments the vertexes with the origin. The partition is made of subdomains associated to extremal arcs of length`(r)and (possibly) to one triple associated to the cluster of parameterh(r). According to (4.1), the former have all area F(r,0), the latter (when present) has areaF(r, h(r)). When r=r2N+1 the partition is regular and
A(r2N+1) = (2N+ 1)F(r2N+1,0).
In all the other cases, namely whenr∈(r2N−1, r2N+1), we have A(r) = (2N−2)F(r,0) +F(r, h(r)).
In the open interval (r2N−1, r2N+1) the functions F(r,0), h(r), and F(r, h(r)) are continuous, therefore A(r) is continuous too. In the limit as r & r2N−1, we have h(r)→0, so that
r&rlim2N−1A(r) = (2N−2)F(r2N−1,0) +F(r2N−1,0) =A(r2N−1);
similarly, when r% r2N+1, we haveh(r)→`(r2N+1) and F(r, `(r)) →3F(r2N+1,0), thus
r%rlim2N+1A(r) = (2N−2)F(r2N+1,0)+3F(r2N+1,0) = (2N+1)F(r2N+1,0) =A(r2N+1).
Therefore,A is continuous in each closed interval [r2N−1, r2N+1]. This concludes the proof of the continuity ofA.
Let us now consider the optimal shapes. We choose the following orientation:
for regular Reuleaux polygons, we take one of the vertexes aligned vertically with the origin, above it; in all the other cases, we choose the pointPk of the cluster (see Definition 3.3) aligned vertically with the origin, below it (see also Figure 1). By con-struction, the position of the vertexes varies continuously with respect to r, so that the optimal shapes vary continuously with respect to the Hausdorff convergence.
Remark4.5. — We could also be interested inthe dual problem, namely tomaximize the inradius among shapes of constant width and fixed area. If we check that the functionr 7→A(r)is strictly increasing (what is clearly true numerically), then an easy argument shows that the domainΩ(r)also solves this dual problem.
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Manuscript received 26th April 2020 accepted 8th February 2021
Antoine Henrot, Université de Lorraine CNRS, IECL F-54000 Nancy, France
E-mail :antoine.henrot@univ-lorraine.fr
Url :http://www.iecl.univ-lorraine.fr/~Antoine.Henrot/
Ilaria Lucardesi, Université de Lorraine CNRS, IECL F-54000 Nancy, France
E-mail :ilaria.lucardesi@univ-lorraine.fr
Url :https://sites.google.com/site/ilarialucardesi/home