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Narrowing the gaps of the missing Blaschke-Santaló diagrams

Bernardo González Merino Área de Matemática Aplicada, Departamento de Ingeniería y Tecnología de Computadores, Facultad de Informática, Universidad de Murcia, 30100-Murcia, Spain bgmerino@um.es
(Date: September 6, 2025)
Abstract.

We solve several new sharp inequalities relating three quantities amongst the area, perimeter, inradius, circumradius, diameter, and minimal width of planar convex bodies. As a consequence, we narrow the missing gaps in each of the missing planar Blaschke-Santaló diagrams. Furthermore, we extend some of those sharp inequalities into higher dimensions, by replacing either the perimeter by the mean width or the area by the volume.

2020 Mathematics Subject Classification. Primary 52A20; Secondary 52A21, 52A40. Partially supported by MICINN Project PID2022-136320NB-I00 Spain.

September 6, 2025

1. Introduction

In 1916 Blaschke [3] raised the question of determining the set of triples given by the volume V(K)V(K), surface area S(K)S(K), and integral mean curvature M(K)M(K) of every set KK from the set of 33-dimensional convex bodies 𝒦3\mathcal{K}^{3}. To do so, he defined the function

h:𝒦3[0,1]2whereh(K):=(4πS(K)M(K)2,48π2V(K)M(K)3),h:\mathcal{K}^{3}\rightarrow[0,1]^{2}\quad\text{where}\quad h(K):=\left(\frac{4\pi S(K)}{M(K)^{2}},\frac{48\pi^{2}V(K)}{M(K)^{3}}\right),

and studied the set of images h(𝒦3)h(\mathcal{K}^{3}) of hh, which is known as the Blaschke diagram. Despite recent efforts (see [29] or [33]) this diagram still today remains unsolved.

Years later, Santaló [34] proposed the study of possible triples for the geometric quantities area AA, perimeter pp, inradius rr, circumradius RR, diameter DD, and minimum width ww of a 22-dimensional convex body. As a result, he already solved the cases (A,p,w)(A,p,w), (A,p,r)(A,p,r), (A,p,R)(A,p,R), (A,D,w)(A,D,w), (p,D,w)(p,D,w), and (R,r,D)(R,r,D). Later on Hernández Cifre and Segura Gomis in [19] and [23] concluded the quest to describe the diagrams (R,r,w)(R,r,w), (R,D,w)(R,D,w), (r,D,w)(r,D,w), Hernández Cifre completely described in [20] the diagrams (A,D,R)(A,D,R) and (p,D,R)(p,D,R), Böröczky Jr., Hernández Cifre, and Salinas described the diagrams (A,r,R)(A,r,R) and (p,r,R)(p,r,R) in [5], and very recently Delyon, Henrot, and Privat completely described in [12] the diagram (A,r,D)(A,r,D).

Those functionals have been also considered in other diagrams. For instance, it has been described the case (R,r,D,w)(R,r,D,w) in [6] as well as an almost complete description by Ting and Keller of (A,p,D,w)(A,p,D,w) in [38]. Furthermore, Hernández Cifre et al. described in [21] diagrams for the inradius, circumradius, and volume and surface area for nn-dimensional centrally symmetric convex bodies. In [7] the authors described detailed (R,r,D)(R,r,D) diagrams when the functionals are measured with respect to other balls different from the Euclidean ball, such as the triangle, square, pentagon, or hexagon, or a general convex body. Very recently in [10] the authors studied (R,r,D)(R,r,D) diagrams for other notions of diameters.

More diagrams have been recently studied when introducing other functionals. Ftouhi and Lamboley [16] have described the diagram for the area, perimeter, and the first Dirichlet eigenvalue, Lucardesi and Zucco [28] described the diagram for the area, the torsional ridigity, and the first Dirichlet eigenvalue (see also Buttazzo and Pratelli [11] for more information on this diagram), Ftouhi and Henrot [15] studied the diagram for the area, the first Dirichtel eigenvalue, and the first non-trivial Laplacian-Neumann eigenvalue, and also Ftouhi [14] computed the diagram of the area, perimeter, and the Cheeger constant. Very recently Gastaldello, Henrot, and Lucardesi in [17] have studied the diagram of area, perimeter, and moment of inertia.

The aim of this paper is to solve several new inequalities relating three of the six functionals proposed by Santaló, and to clarify what is known up to date in the matter. This is why we state and prove each new result in the corresponding section. Indeed, we devote a section to each of the six unsolved diagrams, with the exception of (A,r,w)(A,r,w) and (p,r,w)(p,r,w) which use a common one.

2. Previous inequalities

Let 𝒦n\mathcal{K}^{n} be the set of nn-dimensional convex bodies, i.e., convex and compact sets in n\mathbb{R}^{n} different from points.

For every x,ynx,y\in\mathbb{R}^{n}, x=(x1,x2,,xn)x=(x_{1},x_{2},\dots,x_{n}), y=(y1,y2,,yn)y=(y_{1},y_{2},\dots,y_{n}), let x,y:=x1y1+x2y2++xnyn\langle x,y\rangle:=x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n} be the scalar product of xx and yy. Moreover, let x:=x,x\|x\|:=\sqrt{\langle x,x\rangle} be the Euclidean norm of xx. Furthermore, let e1,e2,,ene_{1},e_{2},\dots,e_{n} be the vectors of the canonical base of n\mathbb{R}^{n}.

Let K𝒦nK\in\mathcal{K}^{n}. The circumradius (resp. inradius) R(K)R(K) (resp. r(K)r(K)) of KK is the smallest (resp. largest) radius of an Euclidean ball containing KK (resp. contained in KK). The diameter D(K)D(K) of KK is the largest Euclidean length between two points contained in KK. The (minimum) width is the minimum distance between two parallel hyperplanes containing KK between them. For every K𝒦2K\in\mathcal{K}^{2}, let p(K)p(K) be the perimeter of KK. For every K𝒦nK\in\mathcal{K}^{n}, let V(K)V(K) be its volume or Lebesgue measure, and if n=2n=2, realize that VV coincides with the usual area AA.

These functionals are monotonically increasing with respect to inclusion, i.e., if K,C𝒦nK,C\in\mathcal{K}^{n}, then f(K)f(C)f(K)\leq f(C) for f{R,r,D,w,V}f\in\{R,r,D,w,V\} and if n=2n=2 then f(K)f(C)f(K)\leq f(C) for f=pf=p too. Moreover, if f{p,R,r,D,w}f\in\{p,R,r,D,w\} then ff is 11-homogeneous, i.e. f(λK)=λf(K)f(\lambda K)=\lambda f(K) and λ0\lambda\geq 0. In the case of the volume, it is nn-homogenous, i.e. V(λK)=λnV(K)V(\lambda K)=\lambda^{n}V(K), for every K𝒦nK\in\mathcal{K}^{n} and λ0\lambda\geq 0, and thus AA is 22-homogeneous.

Some remarkable examples of planar convex bodies are the Euclidean unit ball 𝔹\mathbb{B}, the equilateral triangle of unit circumradius 𝕋\mathbb{T}, the line segment of edge length 22 𝕃\mathbb{L}, and the Reuleaux triangle of unit circumradius 𝕋\mathbb{RT}. A body K𝒦2K\in\mathcal{K}^{2} is of constant width if w(K)=D(K)w(K)=D(K) (see [30] for further details). In nn-dimensional space n\mathbb{R}^{n}, let 𝔹n\mathbb{B}_{n} be the nn-dimensional Euclidean unit ball, thus having that 𝔹2=𝔹\mathbb{B}_{2}=\mathbb{B}.

Since this is eminently a paper devoted to planar sets, when we refer to some result, we will directly specify its 22-dimensional version. Moreover, for the references to most of the known inequalities we will refer to [1] as well as the references therein. However, we state here the inequalities that we need further on.

Let K𝒦2K\in\mathcal{K}^{2}. The isoperimetric inequality (see [31]) states that

(1) 4πA(K)p(K)2,4\pi A(K)\leq p(K)^{2},

with equality if and only if K=𝔹K=\mathbb{B}.

Pál’s inequality (see [32]) provides a lower bound of AA in terms of ww, namely

(2) w(K)23A(K),w(K)^{2}\leq\sqrt{3}A(K),

with equality if and only if K=𝕋K=\mathbb{T}. If KK is moreover assumed to be of constant width, then the inequality strengthens onto

(3) (π3)w(K)22A(K),(\pi-\sqrt{3})w(K)^{2}\leq 2A(K),

known as the Blaschke-Lebesgue inequality (see [4] or [27]), with equality if and only if K=𝕋K=\mathbb{RT}.

Further trivial inequalities between the area and the circumradius and inradius are

πr(K)2A(K)πR(K)2,\pi r(K)^{2}\leq A(K)\leq\pi R(K)^{2},

with equality if and only if K=𝔹K=\mathbb{B}.

The perimeter and the diameter relate as

(4) 2D(K)p(K)πD(K),2D(K)\leq p(K)\leq\pi D(K),

where the left inequality is trivial and gets equal when K=𝕃K=\mathbb{L}, and the right one is a consequence of Barbier [2] and it becomes equality when KK is a constant width set.

The perimeter and the inradius fulfill the trivial inequality 2πr(K)p(K)2\pi r(K)\leq p(K) with equality for K=𝔹K=\mathbb{B}. The perimeter and the circumradius fulfill

(5) 4R(K)p(K)2πR(K),4R(K)\leq p(K)\leq 2\pi R(K),

where the left inequality attains equality when K=𝕃K=\mathbb{L}, and the right inequality gets equal when K=𝔹K=\mathbb{B}.

The relation between the perimeter and the width is described by

(6) πw(K)p(K)\pi w(K)\leq p(K)

as a consequence of Cauchy’s formula (see [35]), with equality if and only if KK is of constant width.

In [5] it was proven the optimal lower bound for the perimeter in terms of the inradius and the circumradius, namely:

(7) p(K)4(R(K)2r(K)2+r(K)arcsin(r(K)R(K))).p(K)\geq 4\left(\sqrt{R(K)^{2}-r(K)^{2}}+r(K)\arcsin\left(\frac{r(K)}{R(K)}\right)\right).

Moreover, equality holds if and only if KK is the convex hull of a point, its mirrored with the origin, and an euclidean ball centered at the origin.

In [25] the author derived the optimal upper bound of the perimeter in terms of the diameter and the width

(8) p(K)2D(K)2w(K)2+2D(K)arcsin(w(K)D(K)).p(K)\leq 2\sqrt{D(K)^{2}-w(K)^{2}}+2D(K)\arcsin(\frac{w(K)}{D(K)}).

Moreover, equality holds for the intersection of the Euclidean ball centered at the origin with a halfspace and its mirrored with respect to the origin.

The quotient between the circumradius and the diameter is described by Jung’s inequality [24]

(9) 3R(K)D(K),\sqrt{3}R(K)\leq D(K),

with equality if and only if KK contains an equilateral triangle of circumradius R(K)R(K) and diameter D(K)D(K).

The relation between the inradius and the width is described by

(10) 2r(K)w(K)3r(K),2r(K)\leq w(K)\leq 3r(K),

where the left side attains equality if and only if K=𝔹K=\mathbb{B} and the right side, known as Steinhagen’s inequality [37], becomes equality if and only if K=𝕋K=\mathbb{T}.

The relation between the circumradius and the width is described by

w(K)2R(K),w(K)\leq 2R(K),

with equality if and only if K=𝔹K=\mathbb{B}.

Finally, if KK is of constant width, it is known that 2R(K)(3+1)r(K)2R(K)\leq(\sqrt{3}+1)r(K) with equality if and only if K=𝕋K=\mathbb{RT}. This last statement is a consequence of (9) and the fact that constant width sets attain equality in the concentricity inequalities

(11) w(K)r(K)+R(K)D(K)w(K)\leq r(K)+R(K)\leq D(K)

(see [8] for more information on these inequalities). Moreover, precisely in the case of constant width sets, one can obtain the sharpened Steinhagen’s inequality

(12) 2w(K)(3+3)r(K)2w(K)\leq(3+\sqrt{3})r(K)

with equality if and only if K=𝕋K=\mathbb{RT}.

Let K𝒦nK\in\mathcal{K}^{n}. Given a hyperplane LnL\subset\mathbb{R}^{n}, we say that KK^{\prime} is the Steiner symmetrization of KK with respect to LL if for every xLx\in L, (x+L)K(x+L^{\bot})\cap K^{\prime} is a segment centered around xx, and it has the same length than (x+L)K(x+L^{\bot})\cap K. It is well known that D(K)D(K)D(K^{\prime})\leq D(K), R(K)R(K)R(K^{\prime})\leq R(K), r(K)r(K)r(K)\leq r(K^{\prime}), V(K)=V(K)V(K^{\prime})=V(K), and if n=2n=2 then A(K)=A(K)A(K^{\prime})=A(K) and p(K)p(K)p(K^{\prime})\leq p(K) (see [35]).

For every XnX\subset\mathbb{R}^{n}, let aff(X)\mathrm{aff}(X), lin(X)\mathrm{lin}(X), and conv(X)\mathrm{conv}(X) be the affine, linear, and convex hull of XX, respectively. Given x,ynx,y\in\mathbb{R}^{n}, we let [x,y]=conv({x,y})[x,y]=\mathrm{conv}(\{x,y\}) be the line segment with endpoints xx and yy. Let K\partial K to be the boundary of KK.

The circumball is characterized by some touching conditions of boundary points (see for instance [9]).

Proposition 2.1.

Let K𝒦nK\in\mathcal{K}^{n} with K𝔹nK\subset\mathbb{B}_{n}. The following are equivalent:

  1. (1)

    R(K)=1R(K)=1.

  2. (2)

    There exist xiK𝔹nx_{i}\in\partial K\cap\partial\mathbb{B}_{n}, i=1,,ji=1,\dots,j, j{2,,n+1}j\in\{2,\dots,n+1\}, fulfilling the property that 0conv({x1,,xj})0\in\mathrm{conv}(\{x_{1},\dots,x_{j}\}).

An analogous result applies to the inradius, namely, if 𝔹nK\mathbb{B}_{n}\subset K, then r(K)=1r(K)=1 if and only if there exist xiK𝔹nx_{i}\in\partial K\cap\partial\mathbb{B}_{n}, i=1,,ji=1,\dots,j, j{2,,n+1}j\in\{2,\dots,n+1\}, such that 0conv({x1,,xj})0\in\mathrm{conv}(\{x_{1},\dots,x_{j}\}). In particular, the points xix_{i} coincide with outer normals of KK and 𝔹n\mathbb{B}_{n}, and thus, whenever n=2n=2 and j=3j=3, the intersection of the halfspaces determined by the orthogonal lines to each xix_{i} containing xix_{i}, i=1,2,3i=1,2,3, determines a triangle TT with the property that

(13) 𝔹KT and r(T)=1.\mathbb{B}\subset K\subset T\quad\text{ and }\quad r(T)=1.

3. (A,p,D)

Let us consider the diagram fApD(𝒦2)f_{ApD}(\mathcal{K}^{2}) where

fApD:𝒦2[0,)2withfApD(K):=(p(K)D(K),A(K)D(K)2)f_{ApD}:\mathcal{K}^{2}\rightarrow[0,\infty)^{2}\quad\text{with}\quad f_{ApD}(K):=\left(\frac{p(K)}{D(K)},\frac{A(K)}{D(K)^{2}}\right)

(see Figure 1).

We expose the best known inequalities relating the three quantities AA, pp, and DD for K𝒦2K\in\mathcal{K}^{2}.

First of all, notice that (4) induces a linear boundary of the diagram fApD(𝒦2)f_{ApD}(\mathcal{K}^{2}). Indeed, KK fulfills (4) with equality if and only if KK is of constant width. In particular, from (1) and (3) we know that if KK is of constant width with w(K)=2w(K)=2, then 2(π3)A(K)4π2(\pi-\sqrt{3})\leq A(K)\leq 4\pi, with K=𝕋K=\mathbb{RT} and K=𝔹K=\mathbb{B} attaining the lowest and highest values (see Figure 1).

Second, Kubota [25] proved that

(14) 8φA(K)p(K)(p(K)2D(K)cosφ),8\varphi A(K)\leq p(K)(p(K)-2D(K)\cos\varphi),

where φ\varphi is defined to be solution of 2φD(K)=p(K)sinφ2\varphi D(K)=p(K)\sin\varphi, with equality for KK being a symmetric lenses, i.e., the intersection of two Euclidean balls of the same diameter. It is worth mentioning that (14) implies a strengthening of the isoperimetric inequality (1) (see Figure 1111Here and on each other diagram, we represent valid inequalities by lines. In case of continuous lines, we refer to real boundaries of the diagram; in case of dashed lines, not. If we draw KK on a diagram, it coincides with the value f(K)f(K) of its image through the corresponding diagram.).

Third, Kubota [25, 26] also proved that if 2D(K)p(K)3D(K)2D(K)\leq p(K)\leq 3D(K) then

(15) 4A(K)(p(K)2D(K))4p(K)D(K)p(K)24A(K)\geq(p(K)-2D(K))\sqrt{4p(K)D(K)-p(K)^{2}}

with equality for isosceles triangles with two equal longer edges, having as extreme cases K=𝕃K=\mathbb{L} and K=𝕋K=\mathbb{T}. Moreover, in the case of 3D(K)p(K)πD(K)3D(K)\leq p(K)\leq\pi D(K), he also proved the inequality

(16) 4A(K)3D(K)(p(K)2D(K)),4A(K)\geq\sqrt{3}D(K)(p(K)-2D(K)),

with equality only when K=𝕋K=\mathbb{T}.

Refer to caption
Figure 1. What we know about the diagram (A,p,D)(A,p,D), where x=p/Dx=p/D and y=A/D2y=A/D^{2}. (14) provides the upper boundary, (15) part of the lower boundary, the right-hand side of (4) gives part of the right boundary, and (4) itself and (16) provide bounds on the remaining diagram. On top we display (1).

4. (p,D,r)

Let us consider the diagram fpDr(𝒦2)f_{pDr}(\mathcal{K}^{2}) where

fpDr:𝒦2[0,)2withfpDr(K):=(A(K)D(K)2,p(K)D(K))f_{pDr}:\mathcal{K}^{2}\rightarrow[0,\infty)^{2}\quad\text{with}\quad f_{pDr}(K):=\left(\frac{A(K)}{D(K)^{2}},\frac{p(K)}{D(K)}\right)

(see Figure 2).

It is well known (see [18, (3)]) that

(17) p(K)2D(K)+4r(K),p(K)\leq 2D(K)+4r(K),

with equality for K=𝕃K=\mathbb{L}.

It is also known that the right-hand side of (4) induces part of the upper boundary of the diagram, where constant width sets attain equality, and with extreme points at K=𝕋K=\mathbb{RT} (due to (12)) and K=𝔹K=\mathbb{B}.

We now establish a new inequality, providing the lower boundary of this diagram.

Refer to caption
Figure 2. What we know about the diagram (p,D,r)(p,D,r), where x=r/Dx=r/D and y=p/Dy=p/D. (4) right provides part of the upper boundary, (18) provides the lower boundary, and (4), (19), and (17) imply upper bounds.
Theorem 4.1.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(18) p(K)r(K)4(π2arctan(D(K)24r(K)21)+D(K)24r(K)21).\frac{p(K)}{r(K)}\geq 4\left(\frac{\pi}{2}-\arctan\left(\sqrt{\frac{D(K)^{2}}{4r(K)^{2}}-1}\right)+\sqrt{\frac{D(K)^{2}}{4r(K)^{2}}-1}\right).

Moreover, for every r[0,1]r\in[0,1], there exists Kr𝒦2K_{r}\in\mathcal{K}^{2} for which r(Kr)=rr(K_{r})=r, D(Kr)=2D(K_{r})=2, and KrK_{r} fulfills (18) with equality.

Proof.

After a translation and rescaling of KK, let us assume that [(1,0),(0,1)]K[(-1,0),(0,1)]\subset K with D(K)=2D(K)=2. Moreover, let P+r(K)𝔹KP+r(K)\mathbb{B}\subset K, for some PKP\in K.

Notice that [(1,0),(0,1)](P+r(K)𝔹)[(-1,0),(0,1)]\cap(P+r(K)\mathbb{B})\neq\emptyset. Indeed, assume that r(K)<1r(K)<1 (as otherwise K=𝔹K=\mathbb{B} and the result is true). since P+r(K)𝔹K{(x,y)2:1x1}P+r(K)\mathbb{B}\subset K\subset\{(x,y)\in\mathbb{R}^{2}:-1\leq x\leq 1\}, if we would have that [(1,0),(0,1)](P+r(K)𝔹)=[(-1,0),(0,1)]\cap(P+r(K)\mathbb{B})=\emptyset, then by the convexity of KK we can easily find an Euclidean ball of strictly larger radius than r(K)r(K) within conv([(1,0),(0,1)](P+r(K)𝔹))K\mathrm{conv}([(-1,0),(0,1)]\cup(P+r(K)\mathbb{B}))\subset K, thus contradicting the definition of r(K)r(K).

By convexity, we have that K1:=conv([(1,0),(1,0)](P+r(K)𝔹))KK_{1}:=\mathrm{conv}([(-1,0),(1,0)]\cup(P+r(K)\mathbb{B}))\subset K, and thus by monotonicity of pp then p(K1)p(K)p(K_{1})\leq p(K). Notice that D(K1)=D(K)=2D(K_{1})=D(K)=2 and r(K1)=r(K)r(K_{1})=r(K).

Moreover, since [(1,0),(0,1)](P+r(K)𝔹)[(-1,0),(0,1)]\cap(P+r(K)\mathbb{B})\neq\emptyset, then P+r(K)(±e2)(K1)P+r(K)(\pm e_{2})\in\partial(K_{1}), and thus the lines P+r(K)(±e2)+lin({e1})P+r(K)(\pm e_{2})+\mathrm{lin}(\{e_{1}\}) support K1K_{1}. Let K2K_{2} be the Steiner symmetrization of K1K_{1} with respect to the line lin({e1})\mathrm{lin}(\{e_{1}\}). It is clear that

[(1,0),(1,0)],(P1,0)+r(K)𝔹K2,(P1,0)+r(K)(±e2)K2,[(-1,0),(1,0)],(P_{1},0)+r(K)\mathbb{B}\subset K_{2},\quad(P_{1},0)+r(K)(\pm e_{2})\in\partial K_{2},

where P=(P1,P2)P=(P_{1},P_{2}). Since 2=D(K1)=D([(1,0),(1,0)])D(K2)D(K1)2=D(K_{1})=D([(-1,0),(1,0)])\leq D(K_{2})\leq D(K_{1}), then D(K2)=2D(K_{2})=2. Moreover, it is also clear that every vertical line intersecting K1K_{1} is a line segment of length at most 2r(K1)2r(K_{1}), and so r(K2)r(K1)=r(K)r(K_{2})\leq r(K_{1})=r(K). This, together with (P1,0)+r(K)𝔹K2(P_{1},0)+r(K)\mathbb{B}\subset K_{2} implies that r(K2)=r(K)=r(K1)r(K_{2})=r(K)=r(K_{1}). By properties of the Steiner symmetrization we know that p(K2)p(K1)p(K_{2})\leq p(K_{1}).

Now again, we do another Steiner symmetrization K3K_{3} of K2K_{2}, this time with respect to the line lin(e2)\mathrm{lin}(e_{2}). We still have all previous properties inherited from K1K_{1} on K2K_{2}. It is easy and almost analogous to check that D(K3)=D([(1,0),(1,0)])=2D(K_{3})=D([(-1,0),(1,0)])=2, r(K3)=r(K2)=r(K)r(K_{3})=r(K_{2})=r(K), and that p(K3)p(K2)p(K1)p(K)p(K_{3})\leq p(K_{2})\leq p(K_{1})\leq p(K). Finally, letting K4:=conv([(1,0),(1,0)](r(K)𝔹))K3K_{4}:=\mathrm{conv}([(-1,0),(1,0)]\cup(r(K)\mathbb{B}))\subset K_{3}, we can clearly see that

r(K4)=r(K),D(K4)=D(K)=2,andp(K)p(K4).r(K_{4})=r(K),\quad D(K_{4})=D(K)=2,\quad\text{and}\quad p(K)\geq p(K_{4}).

Let us now compute p(K4)p(K_{4}), for given values D(K4)=2D(K_{4})=2 and r(K4)=r[0,1]r(K_{4})=r\in[0,1]. To do so, let us compute the only line with negative slope supporting r𝔹r\mathbb{B} and passing through (1,0)(1,0). Let y=m(x1)y=m(x-1), m<0m<0, be this line, such that it is tangent to r𝔹r\mathbb{B}, i.e. such that

{y=m(x1)x2+y2=r2\left\{\begin{array}[]{c}y=m(x-1)\\ x^{2}+y^{2}=r^{2}\end{array}\right.

has a unique solution. Substituting yy onto the second equations gives x2+m2(x1)2=r2x^{2}+m^{2}(x-1)^{2}=r^{2}, i.e. (1+m2)x22m2x+m2r2=0(1+m^{2})x^{2}-2m^{2}x+m^{2}-r^{2}=0. This equation has a unique solution - root - if and only if m4(1+m2)(m2r2)=0m^{4}-(1+m^{2})(m^{2}-r^{2})=0, i.e. if and only if m=r1r2m=-\frac{r}{\sqrt{1-r^{2}}}. In this case, the intersecting point between the line and the circumpherence has coordinates

x0=m21+m2andy0=m1+m2.x_{0}=\frac{m^{2}}{1+m^{2}}\quad\text{and}\quad y_{0}=\frac{-m}{1+m^{2}}.

Since p(K4)p(K_{4}) is symmetric with respect to lin(ei)\mathrm{lin}(e_{i}), i=1,2i=1,2, its perimeter equals four times its perimeter on the first orthant. Since the angle of (x0,y0)(x_{0},y_{0}) with the horizontal line equals arctan(y0/x0)\arctan(y_{0}/x_{0}), we can conclude that

p(K)p(K4)=4(r(π2arctan(y0/x0))+(x01)2+y02)=4(r(π2arctan(1r2r))+1r2).\begin{split}p(K)\geq p(K_{4})&=4\left(r\left(\frac{\pi}{2}-\arctan(y_{0}/x_{0})\right)+\sqrt{(x_{0}-1)^{2}+y_{0}^{2}}\right)\\ &=4\left(r\left(\frac{\pi}{2}-\arctan\left(\frac{\sqrt{1-r^{2}}}{r}\right)\right)+\sqrt{1-r^{2}}\right).\end{split}

In order to derive (18), we simply apply the above inequality to 2K/D(K)2K/D(K) (which has diameter 22) and use the 11-homogeneity of pp and rr.

Moreover, it is clear from the construction that every set K=conv([e1,e1](r𝔹))K=\mathrm{conv}([-e_{1},e_{1}]\cup(r\mathbb{B})), for every r[0,1]r\in[0,1], attains equality in (18). ∎

We now derive an easy non tight upper bound of the diagram as a consequence of (8).

Corollary 4.2.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(19) p(K)4D(K)29r(K)2+2D(K)arcsin(3r(K)D(K)).p(K)\leq\sqrt{4D(K)^{2}-9r(K)^{2}}+2D(K)\arcsin\left(\frac{3r(K)}{D(K)}\right).
Proof.

After a suitable dilatation, let us assume that D(K)=2D(K)=2, w:=w(K)w:=w(K), and r:=r(K)r:=r(K). Using (8) then

p(K)24w2+4arcsin(w2)=:f(w).p(K)\leq 2\sqrt{4-w^{2}}+4\arcsin\left(\frac{w}{2}\right)=:f(w).

Since f(w)f(w) is clearly increasing on w[0,2]w\in[0,2], using (10) we can derive that

p(K)249r2+4arcsin(3r2).p(K)\leq 2\sqrt{4-9r^{2}}+4\arcsin\left(\frac{3r}{2}\right).

In order to obtain the inequality, we simply need to replace KK by 2K/D(K)2K/D(K) and use the 11-homogeneity of pp and rr. ∎

5. (p,R,w)

Let us consider the diagram fpRw(𝒦2)f_{pRw}(\mathcal{K}^{2}) where

fpRw:𝒦2[0,)2withfpRw(K):=(w(K)R(K),p(K)R(K))f_{pRw}:\mathcal{K}^{2}\rightarrow[0,\infty)^{2}\quad\text{with}\quad f_{pRw}(K):=\left(\frac{w(K)}{R(K)},\frac{p(K)}{R(K)}\right)

(see Figure 3).

The inequality (6) induces part of the right boundary of the diagram. Remember that KK belongs to this boundary if and only if KK is of constant width. One of the extreme points is K=𝔹K=\mathbb{B}. On the other hand, if KK is of constant width, i.e. D(K)=w(K)D(K)=w(K), together with (9), implies that 3R(K)w(K)\sqrt{3}R(K)\leq w(K), with equality when K=𝕋K=\mathbb{RT}. Thus, this last set is the other extreme point of this boundary.

Refer to caption
Figure 3. What we know about the diagram (p,R,w)(p,R,w), where x=w/Rx=w/R and y=p/Ry=p/R. (6) induces part of the right boundary, (20) induces the upper boundary, and (6) and (21) provide bounds of the diagram. On top and bottom we show (5).

We now show a new inequality which induces the upper boundary of the diagram.

Theorem 5.1.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(20) p(K)2R(K)π+21w(K)24R(K)22arccos(w(K)2R(K)).\frac{p(K)}{2R(K)}\leq\pi+2\sqrt{1-\frac{w(K)^{2}}{4R(K)^{2}}}-2\arccos\left(\frac{w(K)}{2R(K)}\right).

Moreover, for every w[0,2]w\in[0,2], there exists Kw𝒦2K_{w}\in\mathcal{K}^{2} such that R(Kw)=1R(K_{w})=1, w(Kw)=ww(K_{w})=w, and p(Kw)p(K_{w}) fulfills (20) with equality.

Proof.

After a suitable dilatation and rigid motion, we can assume that R(K)=1R(K)=1 with K𝔹K\subset\mathbb{B}, and such that the lines y=ay=-a and y=way=w-a support KK, where 0awa10\leq a\leq w-a\leq 1 with w:=w(K)w:=w(K), i.e.

if w[0,1] then a[0,w/2]and if w[1,2] then a[w1,w/2].\text{if }w\in[0,1]\text{ then }a\in[0,w/2]\quad\text{and if }w\in[1,2]\text{ then }a\in[w-1,w/2].

Above, we take the lines containing 0 between them. This is due to Proposition 2.1, since a consequence of R(K)=1R(K)=1 with K𝔹K\subset\mathbb{B} is that 0K0\in K. In particular K𝔹{(x,y)2:aywa}=:KK\subset\mathbb{B}\cap\{(x,y)\in\mathbb{R}^{2}:-a\leq y\leq w-a\}=:K^{\prime}, and thus p(K)p(K)p(K)\leq p(K^{\prime}). If we denote by l1l_{1} the arc of 𝕊1\mathbb{S}^{1} in the region yway\geq w-a, l2l_{2} the arc of 𝔹\partial\mathbb{B} in the region yay\leq-a, l3:=𝔹(l1l2)l_{3}:=\partial\mathbb{B}\setminus(l_{1}\cup l_{2}), l4l_{4} the line-segment of y=way=w-a within 𝔹\mathbb{B}, and l5l_{5} the line-segment of y=ay=-a within 𝔹\mathbb{B}, then it is clear that

p(K)=l3+l4+l5=2π2arccos(wa)2arccos(a)+1(wa)2+1a2=:fw(a).\begin{split}p(K^{\prime})&=l_{3}+l_{4}+l_{5}\\ &=2\pi-2\arccos(w-a)-2\arccos(a)+\sqrt{1-(w-a)^{2}}+\sqrt{1-a^{2}}=:f_{w}(a).\end{split}

Notice that

fw(a)=2(wa11(wa)21a1a2).f^{\prime}_{w}(a)=2\left(\frac{w-a-1}{\sqrt{1-(w-a)^{2}}}-\frac{1-a}{\sqrt{1-a^{2}}}\right).

We aim to show that the above expression is nonnegative whenever 0awa10\leq a\leq w-a\leq 1 (which would imply that fw(a)f_{w}(a) is nondecreasing on aa). To do so, notice that it would then be enough to show that the function g(x):=1x1x2g(x):=\frac{1-x}{\sqrt{1-x^{2}}} is decreasing for every x[0,1]x\in[0,1]. Since

g(x)=x12(1x2)1x20g^{\prime}(x)=\frac{x-1}{2(1-x^{2})\sqrt{1-x^{2}}}\leq 0

for every x[0,1]x\in[0,1], the assertion holds. Therefore

p(K)=fw(a)fw(w2)=2(π2arccos(w2)+21(w2)2).p(K^{\prime})=f_{w}(a)\leq f_{w}\left(\frac{w}{2}\right)=2\left(\pi-2\arccos\left(\frac{w}{2}\right)+2\sqrt{1-\left(\frac{w}{2}\right)^{2}}\right).

In order to conclude the proof of the inequality, we simply replace KK by K/R(K)K/R(K) and use the homogeneity of the functionals pp and ww.

Evidently, equality holds for every K=𝔹{(x,y)2:|y|w/2}K=\mathbb{B}\cap\{(x,y)\in\mathbb{R}^{2}:|y|\leq w/2\}, for every w[0,2]w\in[0,2]. ∎

We can easily derived lower bounds, for instance, using (10) in combination with (7).

Corollary 5.2.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(21) p(K)4(R(K)2w(K)29+w(K)3arcsin(w(K)3R(K))).p(K)\geq 4\left(\sqrt{R(K)^{2}-\frac{w(K)^{2}}{9}}+\frac{w(K)}{3}\arcsin\left(\frac{w(K)}{3R(K)}\right)\right).

Moreover, equality holds if K=𝕃K=\mathbb{L}.

Proof.

Let us suppose that R(K)=1R(K)=1, r:=r(K)r:=r(K), and w:=w(K)w:=w(K). Notice that f(r):=1r2+rarcsinrf(r):=\sqrt{1-r^{2}}+r\arcsin r is an increasing function in r[0,1]r\in[0,1] since f(r)=arcsinrf^{\prime}(r)=\arcsin r. Hence, using (10) in combination with (7) we get that

p(K)4(1r2+rarcsinr)4(1w29+w3arcsin(w3)).p(K)\geq 4\left(\sqrt{1-r^{2}}+r\arcsin r\right)\geq 4\left(\sqrt{1-\frac{w^{2}}{9}}+\frac{w}{3}\arcsin\left(\frac{w}{3}\right)\right).

We then obtain the desired result when applying the above inequality to the set K/R(K)K/R(K) and using the 11-homogeneity of pp and ww.

Evidently, when K=𝕃K=\mathbb{L} we attain equality in (21). ∎

6. (A,R,w)

Let us consider the diagram fARw(𝒦2)f_{ARw}(\mathcal{K}^{2}) where

fARw:𝒦2[0,)2withfARw(K):=(w(K)R(K),A(K)R(K)2)f_{ARw}:\mathcal{K}^{2}\rightarrow[0,\infty)^{2}\quad\text{with}\quad f_{ARw}(K):=\left(\frac{w(K)}{R(K)},\frac{A(K)}{R(K)^{2}}\right)

(see Figure 4).

Apparently, the only two known inequalities relating this triple are

(22) 32R(K)w(K)A(K)2R(K)w(K)\frac{\sqrt{3}}{2}R(K)w(K)\leq A(K)\leq 2R(K)w(K)

(see [18]) with both inequalities having equality if K=𝕃K=\mathbb{L}, and the left-hand side also attaining equality at K=𝕋K=\mathbb{T}.

Refer to caption
Figure 4. What we know about the (A,R,w)(A,R,w) diagram, where x=w/Rx=w/R and y=A/R2y=A/R^{2}. (23) provides the upper boundary, (24) and the left-hand side of (22) provide lower bounds for the diagram. Over (23) we see the right-hand side of (22).

The first inequality we prove induces the upper boundary of the diagram.

Theorem 6.1.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(23) A(K)R(K)2(π2arccos(w(K)2R(K))+w(K)R(K)1(w(K)2R(K))2).A(K)\leq R(K)^{2}\left(\pi-2\arccos\left(\frac{w(K)}{2R(K)}\right)+\frac{w(K)}{R(K)}\sqrt{1-\left(\frac{w(K)}{2R(K)}\right)^{2}}\right).

Moreover, for every w[0,2]w\in[0,2], there exists Kw𝒦2K_{w}\in\mathcal{K}^{2} such that R(Kw)=1R(K_{w})=1, w(Kw)=ww(K_{w})=w, and A(Kw)A(K_{w}) fulfill (23) with equality.

Proof.

After a translation and dilatation, let us assume that K𝔹K\subset\mathbb{B} with R(K)=1R(K)=1. Moreover, after a suitable rotation, let us assume that the lines supporting KK with width w:=w(K)w:=w(K) are given by the equations y=way=w-a and y=ay=-a, for some a[0,1]a\in[0,1] such that awaa\leq w-a, i.e.

if w[0,1] then a[0,w/2] and if w[1,2] then a[w1,w/2].\text{if }w\in[0,1]\text{ then }a\in[0,w/2]\text{ and if }w\in[1,2]\text{ then }a\in[w-1,w/2].

Above, we take the lines containing 0 between them. This is due to Proposition 2.1, since a consequence of R(K)=1R(K)=1 with K𝔹K\subset\mathbb{B} is that 0K0\in K. In particular, we have that

K𝔹{(x,y)2:aywa}=:K.K\subset\mathbb{B}\cap\{(x,y)\in\mathbb{R}^{2}:-a\leq y\leq w-a\}=:K^{\prime}.

Thus A(K)A(K)A(K)\leq A(K^{\prime}). Notice that R(K)=R(K)=1R(K^{\prime})=R(K)=1 and w(K)=w(K)=ww(K^{\prime})=w(K)=w. Now let us compute A(K)A(K^{\prime}). Notice that KK^{\prime} is the union of four triangles and two circular sectors as in Figure 5.

Refer to caption
Figure 5. KK is contained between the lines and within 𝔹\mathbb{B}.

The areas of the triangles are (wa)1(wa)2/2(w-a)\sqrt{1-(w-a)^{2}}/2 and a1a2/2a\sqrt{1-a^{2}}/2 (twice each). The area of each of the circular sectors is (πarccos(wa)arccos(a))/2(\pi-\arccos(w-a)-\arccos(a))/2. Thus

A(K)=(πarccos(wa)arccos(a))+(wa)1(wa)2+a1a2=:fw(a).\begin{split}A(K^{\prime})&=(\pi-\arccos(w-a)-\arccos(a))+(w-a)\sqrt{1-(w-a)^{2}}+a\sqrt{1-a^{2}}\\ &=:f_{w}(a).\end{split}

Notice that

fw(a)=2(1+(wa)21(wa)2+1a21a2).f^{\prime}_{w}(a)=2\left(\frac{-1+(w-a)^{2}}{\sqrt{1-(w-a)^{2}}}+\frac{1-a^{2}}{\sqrt{1-a^{2}}}\right).

We aim to show that fw(a)0f^{\prime}_{w}(a)\geq 0 whenever 0awa10\leq a\leq w-a\leq 1. This is equivalent to show

1a2=1a21a21(wa)21(wa)2=1(wa)2,\sqrt{1-a^{2}}=\frac{1-a^{2}}{\sqrt{1-a^{2}}}\geq\frac{1-(w-a)^{2}}{\sqrt{1-(w-a)^{2}}}=\sqrt{1-(w-a)^{2}},

for 0awa10\leq a\leq w-a\leq 1, i.e., that g(x):=1x2g(x):=\sqrt{1-x^{2}} is a decreasing function. Since this last statement is true, we conclude that fw(a)0f^{\prime}_{w}(a)\geq 0, from which

A(K)A(K)=fw(a)fw(w2)=π2arccos(w2)+w1(w2)2.A(K)\leq A(K^{\prime})=f_{w}(a)\leq f_{w}\left(\frac{w}{2}\right)=\pi-2\arccos\left(\frac{w}{2}\right)+w\sqrt{1-\left(\frac{w}{2}\right)^{2}}.

In order to retrieve (23), we only need to replace above KK by K/R(K)K/R(K) and use the 11-homogeneity of ww as well as the 22-homogeneity of AA.

Evidently, every body of the form K=𝔹{(x,y)2:|y|w/2}K=\mathbb{B}\cap\{(x,y)\in\mathbb{R}^{2}:|y|\leq w/2\} attain equality in (23), for every w[0,2]w\in[0,2]. ∎

The second inequality we show relating the area, circumradius, and width is one that gives part of the right bound, partly improving (22) for a certain range of values near K=𝔹K=\mathbb{B}.

Theorem 6.2.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(24) A(K)R(K)2(πarccos(w(K)R(K)1))(w(K)R(K)1)2+(w(K)R(K)1)2w(K)R(K)w(K)2R(K)2.\begin{split}&\frac{A(K)}{R(K)^{2}}\geq\\ &\left(\pi-\arccos\left(\frac{w(K)}{R(K)}-1\right)\right)\left(\frac{w(K)}{R(K)}-1\right)^{2}+\left(\frac{w(K)}{R(K)}-1\right)\sqrt{2\frac{w(K)}{R(K)}-\frac{w(K)^{2}}{R(K)^{2}}}.\end{split}

Moreover, equality holds if K=𝔹K=\mathbb{B}.

Proof.

Let us suppose that K𝔹K\subset\mathbb{B}, R(K)=1R(K)=1, and w:=w(K)w:=w(K). Using the fact that w(K)R(K)+r(K)w(K)\leq R(K)+r(K) (see the left-hand side of (11)), then r:=r(K)w1r:=r(K)\geq w-1. Let t+r𝔹Kt+r\mathbb{B}\subset K for some tKt\in K. Moreover, by Proposition 2.1 let piK𝔹p_{i}\in\partial K\cap\partial\mathbb{B}, i=1,2,3i=1,2,3, be such that 0conv({pi:i=1,2,3})0\in\mathrm{conv}(\{p_{i}:i=1,2,3\}). Hence i=1,2,3λipi=0\sum_{i=1,2,3}\lambda_{i}p_{i}=0, for some λi0\lambda_{i}\geq 0, and thus i=1,2,3λipi,t=0\sum_{i=1,2,3}\lambda_{i}\langle p_{i},t\rangle=0 from which at least one of them, say p1p_{1}, fulfills p1,t0\langle p_{1},t\rangle\leq 0. Using the convexity of KK we get that Kt:=conv({p1}(t+r𝔹))KK^{\prime}_{t}:=\mathrm{conv}(\{p_{1}\}\cup(t+r\mathbb{B}))\subset K and thus A(K)A(Kt)A(K)\geq A(K^{\prime}_{t}). Notice that the distance from p1p_{1} to t+r𝔹t+r\mathbb{B} is at least 2w2-w, which occurs when t=0t=0. Thus A(Kt)A(K0)A(K^{\prime}_{t})\geq A(K^{\prime}_{0}). It is clear that K0K^{\prime}_{0} decomposes in a circular sector of angle 2π2arccos(w1)2\pi-2\arccos(w-1) and radius (w1)(w-1) and two triangles each of area (w1)1(w1)2/2(w-1)\sqrt{1-(w-1)^{2}}/2 (see Figure 6), and thus

A(K)A(K0)=(πarccos(w1))(w1)2+(w1)2ww2.A(K)\geq A(K^{\prime}_{0})=(\pi-\arccos(w-1))(w-1)^{2}+(w-1)\sqrt{2w-w^{2}}.

In order to obtain the result, we simply apply the above inequality to the set K/R(K)K/R(K) and take into account the 11-homogeneity of ww as well as the 22-homogeneity of AA.

Moreover, if we replace KK by 𝔹\mathbb{B}, it is very simple to notice that the inequality above becomes equality.

Refer to caption
Figure 6. KK contains an area at least as large as this figure.

7. (A,r,w) and (p,r,w)

In this last section we consider both remaining diagrams since the techniques to induce both inequalities are completely analogous.

Let us consider the diagrams fArw(𝒦2)f_{Arw}(\mathcal{K}^{2}) as well as fprw(𝒦2)f_{prw}(\mathcal{K}^{2}) where

fArw:𝒦2[0,)2withfArw(K):=(w(K)r(K),A(K)r(K)2)f_{Arw}:\mathcal{K}^{2}\rightarrow[0,\infty)^{2}\quad\text{with}\quad f_{Arw}(K):=\left(\frac{w(K)}{r(K)},\frac{A(K)}{r(K)^{2}}\right)

(see Figure 7) and

fprw:𝒦2[0,)2withfprw(K):=(w(K)r(K),p(K)r(K))f_{prw}:\mathcal{K}^{2}\rightarrow[0,\infty)^{2}\quad\text{with}\quad f_{prw}(K):=\left(\frac{w(K)}{r(K)},\frac{p(K)}{r(K)}\right)

(see Figure 8).

7.1. (A,r,w)(A,r,w)

The left-hand side inequality of (10) induces the left unbounded boundary of the diagram, with extreme point at K=𝔹K=\mathbb{B}. The best known classic upper bounds so far were the inequalities

(25) 4(w(K)2r(K))A(K)w(K)3,3(w(K)2r(K))A(K)w(K)2r(K)4(w(K)-2r(K))A(K)\leq w(K)^{3},\quad\sqrt{3}(w(K)-2r(K))A(K)\leq w(K)^{2}r(K)

(see [36]), where the second one attains equality when K=𝕋K=\mathbb{T}. The best lower bound was so far the inequality (2), with equality if and only if K=𝕋K=\mathbb{T} (see Figure 7).

Refer to caption
Figure 7. What we know about the diagram (A,r,w)(A,r,w), where x=w/rx=w/r and y=A/r2y=A/r^{2}. The left-hand side of (10) gives the left boundary, (27) provides the upper boundary, over it we find both inequalities (25), and (29) gives the best known lower boundary of the diagram.

7.2. (p,r,w)(p,r,w)

The left-hand side inequality of (10) induces the left unbounded boundary of the diagram, with extreme point at K=𝔹K=\mathbb{B}. Moreover, the inequality (6) provides part of the lower bound of the diagram, with extreme points at K=𝔹K=\mathbb{B} and K=𝕋K=\mathbb{RT} (due to (12)). The best known classic upper bound so far was the inequality

(26) 3(w(K)2r(K))p(K)2w(K)2\sqrt{3}(w(K)-2r(K))p(K)\leq 2w(K)^{2}

(see [36]), with equality if and only if K=𝕋K=\mathbb{T}.

Refer to caption
Figure 8. What we know about the diagram (p,r,w)(p,r,w). The left-hand side of (10) gives the left boundary, (28) gives the upper boundary, over it we find (26), (6) gives part of the lower boundary, and (30) provides an estimate of part of the lower boundary, improving upon (6).

The first result we show is a pair of inequalities which induce the upper boundaries of their respective diagrams.

Theorem 7.1.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(27) (w(K)2r(K))4r(K)w(K)A(K)w(K)32r(K)2(w(K)-2r(K))\sqrt{4r(K)-w(K)}A(K)\leq w(K)^{\frac{3}{2}}r(K)^{2}

and

(28) (w(K)2r(K))4r(K)w(K)p(K)2r(K)w(K)w(K).(w(K)-2r(K))\sqrt{4r(K)-w(K)}p(K)\leq 2r(K)w(K)\sqrt{w(K)}.

Moreover, for every w[2,3]w\in[2,3] there exists an isosceles triangle KwK_{w} such that r(Kw)=1r(K_{w})=1, w(Kw)=ww(K_{w})=w, and A(Kw)A(K_{w}) (resp. p(Kw)p(K_{w})) fulfills (27) (resp. (28)) with equality.

Proof.

Let TT be a triangle such that KTK\subset T with r(K)=r(T)r(K)=r(T) (see (13)). Of course we also have that w(K)w(T)w(K)\leq w(T), A(K)A(T)A(K)\leq A(T), and p(K)p(T)p(K)\leq p(T) and hence

w(T)r(T)w(K)r(K),A(T)r(T)2A(K)r(K)2andp(T)r(T)p(K)r(K).\frac{w(T)}{r(T)}\geq\frac{w(K)}{r(K)},\quad\frac{A(T)}{r(T)^{2}}\geq\frac{A(K)}{r(K)^{2}}\quad\text{and}\quad\frac{p(T)}{r(T)}\geq\frac{p(K)}{r(K)}.

Let us assume now that (±D/2,0)T(\pm D/2,0)\in T, with D:=D(T)D:=D(T), and thus the width of TT is attained in the direction e2e_{2}.

In the next step, we will show that replacing the vertex (x0,w(T))(x_{0},w(T)) of TT, x0x_{0}\in\mathbb{R}, not belonging to [(D/2,0),(D/2,0)][(-D/2,0),(D/2,0)] by (x1,w(T))(x_{1},w(T)) such that the new triangle TT^{\prime} is an isosceles triangle with (x1,w(T))(D/2,0)=D\|(x_{1},w(T))-(-D/2,0)\|=D, has the properties r(T)r(T)r(T^{\prime})\leq r(T), A(T)=A(T)A(T^{\prime})=A(T), and p(T)p(T)p(T^{\prime})\geq p(T). To do so, let us consider the points (a,0),(a,0),(b,h)(-a,0),(a,0),(b,h), a,b,h0a,b,h\geq 0, and let

fa,h:=(b,h)(a,0)+(b,h)(a,0)=(b+a)2+h2+(ba)2+h2.\begin{split}f_{a,h}&:=\|(b,h)-(-a,0)\|+\|(b,h)-(a,0)\|\\ &=\sqrt{(b+a)^{2}+h^{2}}+\sqrt{(b-a)^{2}+h^{2}}.\end{split}

Since

fa,h(b)=ba(ba)2+h2+b+a(b+a)2+h2,f_{a,h}^{\prime}(b)=\frac{b-a}{\sqrt{(b-a)^{2}+h^{2}}}+\frac{b+a}{\sqrt{(b+a)^{2}+h^{2}}},

we will show that fa,h(b)0f_{a,h}^{\prime}(b)\geq 0 for every b0b\geq 0, thus showing in particular that p(T)p(T)p(T^{\prime})\geq p(T). Evidently, fa,h(b)0f_{a,h}^{\prime}(b)\geq 0 whenever bab\geq a; thus it only remains to show it when b[0,a]b\in[0,a]. In that case, it is equivalent to show that

1+h2(ab)21+h2(a+b)2,\sqrt{1+\frac{h^{2}}{(a-b)^{2}}}\geq\sqrt{1+\frac{h^{2}}{(a+b)^{2}}},

which holds if and only if (a+b)2(ab)2(a+b)^{2}\geq(a-b)^{2}, which is equivalent to the valid inequality 2b02b\geq 0.

Thus, replacing TT by the isosceles TT^{\prime} would give us p(T)p(T)p(T^{\prime})\geq p(T) (see above), and by Fubini’s theorem that A(T)=A(T)A(T^{\prime})=A(T). Hence, using the inradius, perimeter, and area formula of triangles r(T)p(T)=2A(T)=2A(T)=r(T)p(T)r(T)p(T)=2A(T)=2A(T^{\prime})=r(T^{\prime})p(T^{\prime}) implies that r(T)r(T)r(T^{\prime})\leq r(T). Evidently, we also have that w(T)=w(T)w(T^{\prime})=w(T). Thus

w(T)r(T)w(T)r(T),A(T)r(T)2A(T)r(T)2andp(T)r(T)p(T)r(T).\frac{w(T^{\prime})}{r(T^{\prime})}\geq\frac{w(T)}{r(T)},\quad\frac{A(T^{\prime})}{r(T^{\prime})^{2}}\geq\frac{A(T)}{r(T)^{2}}\quad\text{and}\quad\frac{p(T^{\prime})}{r(T^{\prime})}\geq\frac{p(T)}{r(T)}.

Wrapping up all the above, we have proven that for every KK there exists an isosceles triangle TT^{\prime} with two longer edges with

w(T)r(T)w(K)r(K),A(T)r(T)2A(K)r(K)2andp(T)r(T)p(K)r(K).\frac{w(T^{\prime})}{r(T^{\prime})}\geq\frac{w(K)}{r(K)},\quad\frac{A(T^{\prime})}{r(T^{\prime})^{2}}\geq\frac{A(K)}{r(K)^{2}}\quad\text{and}\quad\frac{p(T^{\prime})}{r(T^{\prime})}\geq\frac{p(K)}{r(K)}.
Refer to caption
Figure 9. An isosceles triangle of one single shorter edge of length 2a2a.

Finally, let α\alpha be one of the two equal angles of TT^{\prime}, and let aa be half of the length of the shorter edge of TT^{\prime} (see Figure 9). Moreover, let us call r:=r(T)=1r:=r(T^{\prime})=1, w:=w(T)w:=w(T^{\prime}), A:=A(T)A:=A(T^{\prime}), and p:=p(T)p:=p(T^{\prime}). Using standard trigonometric arguments we obtain that

sinα=w2aandtanα2=1a.\sin\alpha=\frac{w}{2a}\quad\text{and}\quad\tan\frac{\alpha}{2}=\frac{1}{a}.

From there we easily obtain that

a=w4wandα=2arctan(4ww)a=\sqrt{\frac{w}{4-w}}\quad\text{and}\quad\alpha=2\arctan\left(\sqrt{\frac{4-w}{w}}\right)

(notice that w3w\leq 3 by Steinhagen’s theorem (10)). The base of TT^{\prime} has length 2a2a and the corresponding height equals to h=atanαh=a\tan\alpha.

On the one hand, we have that

A=12(2a)h=(w4w)(w4wtan(2arctan(4ww)))=ww(w2)4w=:g1(w).\begin{split}A&=\frac{1}{2}(2a)h=\left(\sqrt{\frac{w}{4-w}}\right)\left(\sqrt{\frac{w}{4-w}}\tan\left(2\arctan\left(\sqrt{\frac{4-w}{w}}\right)\right)\right)\\ &=\frac{w\sqrt{w}}{(w-2)\sqrt{4-w}}=:g_{1}(w).\end{split}

Since g1(w)=4(w3)w(w2)2(4w)320g_{1}^{\prime}(w)=\frac{4(w-3)\sqrt{w}}{(w-2)^{2}(4-w)^{\frac{3}{2}}}\leq 0 for every w[2,3]w\in[2,3], thus g1g_{1} is non-increasing. Hence, assuming that r(K)=r=1r(K)=r=1 too, we conclude that

A(K)A(T)=w(T)w(T)(w(T)2)4w(T)w(K)w(K)(w(K)2)4w(K).A(K)\leq A(T^{\prime})=\frac{w(T^{\prime})\sqrt{w(T^{\prime})}}{(w(T^{\prime})-2)\sqrt{4-w(T^{\prime})}}\leq\frac{w(K)\sqrt{w(K)}}{(w(K)-2)\sqrt{4-w(K)}}.

The inequality is then retrieved by applying the inequality above to K/r(K)K/r(K) and using the 11-homogeneity of ww as well as the 22-homogeneity of AA.

On the other hand

p=2a+2a2+h2=2(w4w+w4w+w4wtan2(2arctan(4ww)))=2ww(w2)4w=:g2(w).\begin{split}p&=2a+2\sqrt{a^{2}+h^{2}}\\ &=2\left(\sqrt{\frac{w}{4-w}}+\sqrt{\frac{w}{4-w}+\frac{w}{4-w}\tan^{2}\left(2\arctan\left(\sqrt{\frac{4-w}{w}}\right)\right)}\right)\\ &=2\frac{w\sqrt{w}}{(w-2)\sqrt{4-w}}=:g_{2}(w).\end{split}

Since

g2(w)=8(w3)w(w2)2(4w)32,g^{\prime}_{2}(w)=\frac{8(w-3)\sqrt{w}}{(w-2)^{2}(4-w)^{\frac{3}{2}}},

thus g2(w)0g_{2}^{\prime}(w)\leq 0 for every w[2,3]w\in[2,3], hence getting that g2g_{2} is decreasing on w[2,3]w\in[2,3]. Therefore

p(K)p(T)=2w(T)w(T)(w(T)2)4w(T)2w(K)w(K)(w(K)2)4w(K).\begin{split}p(K)\leq p(T^{\prime})=2\frac{w(T^{\prime})\sqrt{w(T^{\prime})}}{(w(T^{\prime})-2)\sqrt{4-w(T^{\prime})}}\leq 2\frac{w(K)\sqrt{w(K)}}{(w(K)-2)\sqrt{4-w(K)}}.\end{split}

In order to obtain the second result, we simply apply the inequality above to K/r(K)K/r(K) and use the 11-homogeneity of the functionals pp and ww.

Moreover, equality holds on each of the inequalities above for every isosceles triangle with two longer equal edges. ∎

We end this section by providing new inequalities that bound from below each of their diagrams. Note that (29) already provides a strenghening of Pál’s inequality (2), whereas (30) improves (6) within a range of values near K=𝕋K=\mathbb{T}.

Theorem 7.2.

Let K𝒦2K\in\mathcal{K}^{2}. Then

(29) A(K)r(K)2π+3(w(K)2r(K)22w(K)r(K)+arcsin(r(K)w(K)r(K))π2)\frac{A(K)}{r(K)^{2}}\geq\pi+3\left(\sqrt{\frac{w(K)^{2}}{r(K)^{2}}-\frac{2w(K)}{r(K)}}+\arcsin\left(\frac{r(K)}{w(K)-r(K)}\right)-\frac{\pi}{2}\right)

and

(30) p(K)6(w(K)22w(K)+r(K)arcsin(r(K)w(K)r(K)))πr(K).p(K)\geq 6\left(\sqrt{w(K)^{2}-2w(K)}+r(K)\arcsin\left(\frac{r(K)}{w(K)-r(K)}\right)\right)-\pi r(K).

Moreover, equality holds in both inequalities when K=𝔹K=\mathbb{B} or K=𝕋K=\mathbb{T}.

Proof.

Let TT be a triangle with KTK\subset T such that r(T)=r(K)r(T)=r(K) (see (13)). Let us suppose after a translation and dilatation that r(K)=1r(K)=1 with 𝔹K\mathbb{B}\subset K. Let LiL_{i} be the edges of TT, i=1,2,3i=1,2,3. Since w(T)w(K)=:ww(T)\geq w(K)=:w, there exists piK𝔹p_{i}\in K\notin\mathbb{B} at distance ρiw(K)\rho_{i}\geq w(K) from LiL_{i}, i=1,2,3i=1,2,3. Notice also that each pip_{i} belongs to one of the three disconnected regions of T𝔹T\setminus\mathbb{B}. Using the convexity of KK, in particular we would have that

K:=i=1,2,3(conv({pi}𝔹)𝔹)𝔹KK^{\prime}:=\bigcup_{i=1,2,3}(\mathrm{conv}(\{p_{i}\}\cup\mathbb{B})\setminus\mathbb{B})\cup\mathbb{B}\subset K

with int(conv({pi}𝔹)int(conv({pj}𝔹)=\mathrm{int}(\mathrm{conv}(\{p_{i}\}\cup\mathbb{B})\cap\mathrm{int}(\mathrm{conv}(\{p_{j}\}\cup\mathbb{B})=\emptyset, for every iji\neq j. Thus A(K)A(K)A(K)\geq A(K^{\prime}) and p(K)p(K)p(K)\geq p(K^{\prime}).

We first compute A(K)A(K^{\prime}). We have that

A(K)=A(𝔹)+i=1,2,3A(conv({pi}𝔹)𝔹).A(K^{\prime})=A(\mathbb{B})+\sum_{i=1,2,3}A(\mathrm{conv}(\{p_{i}\}\cup\mathbb{B})\setminus\mathbb{B}).

Evidently, since the distance from pip_{i} to 𝔹\mathbb{B} is at least w2w-2, we can replace pip_{i} by pip_{i}^{\prime} such that the distance from pip_{i}^{\prime} to 𝔹\mathbb{B} is exactly equal to w2w-2. We now compute A(conv({pi}𝔹)𝔹)A(\mathrm{conv}(\{p_{i}\}\cup\mathbb{B})\setminus\mathbb{B}). To do so, notice that the region can be seen as a rectangular triangle of hipotenuse w2w-2 and leg equal to 11 of common angle α\alpha minus the corresponding ball sector from 𝔹\mathbb{B} of angle α\alpha. On the one hand, the area of the triangle equals (w1)212\frac{\sqrt{(w-1)^{2}-1}}{2}, and on the other hand the area of the ball sector equals α2\frac{\alpha}{2}, where α=π2arcsin(1w1)\alpha=\frac{\pi}{2}-\arcsin\left(\frac{1}{w-1}\right). Adding it all, we would conclude that

A(K)A(K)=π+6((w1)212π2arcsin(1w1)2).A(K)\geq A(K^{\prime})=\pi+6\left(\frac{\sqrt{(w-1)^{2}-1}}{2}-\frac{\frac{\pi}{2}-\arcsin\left(\frac{1}{w-1}\right)}{2}\right).

We now compute p(K)p(K^{\prime}). Using the same ideas than above, it is clear that

p(K)p(K)=6(w1)21+2π2π6α2π=6(w22w+arcsin(1w1))π.\begin{split}p(K)&\geq p(K^{\prime})=6\sqrt{(w-1)^{2}-1}+2\pi\frac{2\pi-6\alpha}{2\pi}\\ &=6\left(\sqrt{w^{2}-2w}+\arcsin\left(\frac{1}{w-1}\right)\right)-\pi.\end{split}

In order to obtain (29) and (30), we simply apply the inequalities above to the set K/r(K)K/r(K) and use the 11-homogeneity of the functionals pp and ww as well as the 22-homogeneity of AA.

Moreover, if we replace KK by 𝔹\mathbb{B} or 𝕋\mathbb{T} then (29) and (30) become equality. ∎

8. Higher dimensions

Some of the inequalities above can be naturally extended to higher dimensions with not much additional effort. To this aim, let 𝕊n1:=𝔹n\mathbb{S}^{n-1}:=\partial\mathbb{B}_{n}. Moreover, let b(K)b(K) be the mean width of K𝒦nK\in\mathcal{K}^{n}, i.e. b(K)=𝕊n1w(K;u)/2𝑑σ(u)b(K)=\int_{\mathbb{S}^{n-1}}w(K;u)/2d\sigma(u), where w(K;u)w(K;u) is the directional width of KK in the direction u𝕊n1u\in\mathbb{S}^{n-1} and σ\sigma is the unitary measure over 𝕊n1\mathbb{S}^{n-1}. Let S(K)S(K) be the surface area measure of KK. Remember that Vn(𝔹n)=πn2/Γ(n/2+1)V_{n}(\mathbb{B}_{n})=\pi^{\frac{n}{2}}/\Gamma(n/2+1), where Γ\Gamma stands for the Gamma function, and that S(𝔹n)=nV(𝔹n)S(\mathbb{B}_{n})=nV(\mathbb{B}_{n}) for every nn\in\mathbb{N}. We will stress the subindex in the volume VnV_{n} when applied to some K𝒦nK\in\mathcal{K}^{n} in case it helps in avoiding confusion.

If LL is an ii-dimensional lineal subspace of n\mathbb{R}^{n}, let PLP_{L} be the orthogonal projection onto LL. Moreover, if lin(K)\mathrm{lin}(K) is 22-dimensional, then let p(K)p(K) be the perimeter of KK measured within lin(K)\mathrm{lin}(K). Notice that for every K𝒦nK\in\mathcal{K}^{n} we have that

(31) b(K)=1S(𝔹n)𝕊n1w(K;u)2𝑑σn1(u)=12πS(𝔹n1)𝕊n20πw(K;cosθv+sinθen)𝑑θ𝑑σn2(v)=12πS(𝔹n1)𝕊n2p(Plin({v,en})(K))𝑑σn2(v).\begin{split}b(K)&=\frac{1}{S(\mathbb{B}_{n})}\int_{\mathbb{S}^{n-1}}\frac{w(K;u)}{2}d\sigma_{n-1}(u)\\ &=\frac{1}{2\pi S(\mathbb{B}_{n-1})}\int_{\mathbb{S}^{n-2}}\int_{0}^{\pi}w(K;\cos\theta v+\sin\theta e_{n})d\theta d\sigma_{n-2}(v)\\ &=\frac{1}{2\pi S(\mathbb{B}_{n-1})}\int_{\mathbb{S}^{n-2}}p(P_{\mathrm{lin}(\{v,e_{n}\})}(K))d\sigma_{n-2}(v).\end{split}

The first result is an extension of (18) when replacing p(K)p(K) by b(K)b(K).

Theorem 8.1.

Let K𝒦nK\in\mathcal{K}^{n}. Then

(32) b(K)2π(r(K)(π2arctan(D(K)2r(K)21))+D(K)2r(K)2).b(K)\geq\frac{2}{\pi}\left(r(K)\left(\frac{\pi}{2}-\arctan\left(\sqrt{\frac{D(K)^{2}}{r(K)^{2}}-1}\right)\right)+\sqrt{D(K)^{2}-r(K)^{2}}\right).

Moreover, for every r[0,1]r\in[0,1], there exists Kr𝒦nK_{r}\in\mathcal{K}^{n} with r(Kr)=rr(K_{r})=r, D(Kr)=2D(K_{r})=2, and such that b(Kr)b(K_{r}) fulfills (32) with equality.

Proof.

After a translation and rescaling of KK, let us assume that [e1,e1]K[-e_{1},e_{1}]\subset K with D(K)=2D(K)=2. Moreover, let P+r(K)𝔹nKP+r(K)\mathbb{B}_{n}\subset K, for some PKP\in K.

It is straighforward to check that [e1,e1](P+r(K)𝔹n)[-e_{1},e_{1}]\cap(P+r(K)\mathbb{B}_{n})\neq\emptyset (as in the proof of (18)). By convexity, we have that K1:=conv([e1,e1](P+r(K)𝔹n))KK_{1}:=\mathrm{conv}([-e_{1},e_{1}]\cup(P+r(K)\mathbb{B}_{n}))\subset K, and thus by monotonicity of bb then b(K1)b(K)b(K_{1})\leq b(K). Notice that D(K1)=D(K)=2D(K_{1})=D(K)=2 and r(K1)=r(K)r(K_{1})=r(K). If we now proceed as in the proof of (18) by applying successive Steiner symmetrizations with respect to the hyperplanes e1,e2,,ene_{1},e_{2},\dots,e_{n}, we would easily conclude that p(K)p(K)p(K)\geq p(K^{\prime}), where K:=conv([e1,e1](r(K)𝔹n))K^{\prime}:=\mathrm{conv}([-e_{1},e_{1}]\cup(r(K)\mathbb{B}_{n})), with D(K)=D(K)=2D(K^{\prime})=D(K)=2 and r(K)=r(K)r(K^{\prime})=r(K).

Now observe that by (31) and (18) applied within lin({v,en})\mathrm{lin}(\{v,e_{n}\}), then

b(K)b(K)=12πS(𝔹n1)𝕊n2p(Klin({v,en}))𝑑σn2(v)2π[r(Klin({v,en}))(π2arctan(1r(Klin({v,en}))2r))+1r(Klin({v,en}))2]=2π[r(K)(π2arctan(1r(K)2r))+1r(K)2],\begin{split}b(K)&\geq b(K^{\prime})=\frac{1}{2\pi S(\mathbb{B}_{n-1})}\int_{\mathbb{S}^{n-2}}p(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))d\sigma_{n-2}(v)\\ &\geq\frac{2}{\pi}\Bigg{[}r(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))\left(\frac{\pi}{2}-\arctan\left(\frac{\sqrt{1-r(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))^{2}}}{r}\right)\right)\\ &+\sqrt{1-r(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))^{2}}\Bigg{]}\\ &=\frac{2}{\pi}\Bigg{[}r(K)\left(\frac{\pi}{2}-\arctan\left(\frac{\sqrt{1-r(K)^{2}}}{r}\right)\right)+\sqrt{1-r(K)^{2}}\Bigg{]},\end{split}

since Plin({v,en})(K)=Klin({v,en})P_{\mathrm{lin}(\{v,e_{n}\})}(K^{\prime})=K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}) and r(Klin({v,en}))=r(K)r(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))=r(K^{\prime}) for every v𝕊n2v\in\mathbb{S}^{n-2}.

The inequality (32) is then retrieved by applying the above inequality to the set K/D(K)K/D(K) and using the 11-homogeneity of the functionals bb and rr.

Evidently, equality holds whenever K=conv([e1,e1](r𝔹n))K=\mathrm{conv}([-e_{1},e_{1}]\cup(r\mathbb{B}_{n})), for every r[0,1]r\in[0,1]. ∎

The next result is an extension of (20) to higher dimensions, replacing p(K)p(K) by b(K)b(K).

Theorem 8.2.

Let K𝒦nK\in\mathcal{K}^{n}. Then

b(K)1πR(K)(π+21w(K)24R(K)22arccos(w(K)2R(K))).b(K)\leq\frac{1}{\pi}R(K)\left(\pi+2\sqrt{1-\frac{w(K)^{2}}{4R(K)^{2}}}-2\arccos\left(\frac{w(K)}{2R(K)}\right)\right).

Moreover, for every w[0,2]w\in[0,2], there exists Kw𝒦nK_{w}\in\mathcal{K}^{n} such that R(Kw)=1R(K_{w})=1, w(Kw)=ww(K_{w})=w, and b(Kw)b(K_{w}) fulfills with equality the inequality above.

Proof.

After a suitable dilatation and rigid motion, we can assume that R(K)=1R(K)=1 with K𝔹nK\subset\mathbb{B}_{n}, and such that the hyperplanes xn=ax_{n}=-a and xn=wax_{n}=w-a support KK, where 0awa10\leq a\leq w-a\leq 1 with w:=w(K)w:=w(K), i.e.

if w[0,1] then a[0,w/2]and if w[1,2] then a[w1,w/2].\text{if }w\in[0,1]\text{ then }a\in[0,w/2]\quad\text{and if }w\in[1,2]\text{ then }a\in[w-1,w/2].

In particular K𝔹n{(x1,,xn)n:axnwa}=:KK\subset\mathbb{B}_{n}\cap\{(x_{1},\dots,x_{n})\in\mathbb{R}^{n}:-a\leq x_{n}\leq w-a\}=:K^{\prime}, and thus b(K)b(K)b(K)\leq b(K^{\prime}). Evidently, R(K)=R(K)=1R(K^{\prime})=R(K)=1 and w(K)=w(K)w(K^{\prime})=w(K).

Using (31) and (20) within lin({v,en})\mathrm{lin}(\{v,e_{n}\}) for every v𝕊n2v\in\mathbb{S}^{n-2} we obtain that

b(K)b(K)=12πS(𝔹n1)𝕊n2p(Klin({v,en}))𝑑σn2(v)1π(π+21w(Klin({v,en}))242arccos(w(Klin({v,en}))2))=1π(π+21w(K)42arccos(w(K)2)).\begin{split}b(K)&\leq b(K^{\prime})=\frac{1}{2\pi S(\mathbb{B}_{n-1})}\int_{\mathbb{S}^{n-2}}p(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))d\sigma_{n-2}(v)\\ &\leq\frac{1}{\pi}\left(\pi+2\sqrt{1-\frac{w(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))^{2}}{4}}-2\arccos\left(\frac{w(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}))}{2}\right)\right)\\ &=\frac{1}{\pi}\left(\pi+2\sqrt{1-\frac{w(K)}{4}}-2\arccos\left(\frac{w(K)}{2}\right)\right).\end{split}

where Plin({v,en})(K)=Klin({v,en})P_{\mathrm{lin}(\{v,e_{n}\})}(K^{\prime})=K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}) and w(K)=w(Klin({v,en})w(K^{\prime})=w(K^{\prime}\cap\mathrm{lin}(\{v,e_{n}\}), for every v𝕊n2v\in\mathbb{S}^{n-2}. In order to obtain the inequality, one simply needs to replace above KK by K/R(K)K/R(K) and use the 11-homogeneity of the functionals bb and ww.

Evidently, equality holds whenever K=𝔹n{(x1,,xn)n:|xn|w/2}K=\mathbb{B}_{n}\cap\{(x_{1},\dots,x_{n})\in\mathbb{R}^{n}:|x_{n}|\leq w/2\}, for every w[0,2]w\in[0,2]. ∎

Let us remember a particular instance of the Eulerian hypergeometric function

H(12,1n2;32;a2)=1a0a(1t2)n12𝑑tH\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};a^{2}\right)=\frac{1}{a}\int_{0}^{a}(1-t^{2})^{\frac{n-1}{2}}dt

for every nn\in\mathbb{N} and a(0,1]a\in(0,1] (see [13]).

The following result is an extension of (23) to higher dimensions when replacing A(K)A(K) by V(K)V(K).

Theorem 8.3.

Let K𝒦nK\in\mathcal{K}^{n}. Then

(33) V(K)πn12Γ(n+12)w(K)R(K)n1H(12,1n2;32;w(K)24R(K)2).V(K)\leq\frac{\pi^{\frac{n-1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}w(K)R(K)^{n-1}H\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\frac{w(K)^{2}}{4R(K)^{2}}\right).

Moreover, for every w[0,2]w\in[0,2], there exists Kw𝒦nK_{w}\in\mathcal{K}^{n} such that R(Kw)=1R(K_{w})=1, w(Kw)=ww(K_{w})=w, and V(Kw)V(K_{w}) fulfills (33) with equality.

Proof.

After a suitable dilatation and rigid motion, we can assume that R(K)=1R(K)=1 with K𝔹K\subset\mathbb{B}, and such that the hyperplanes xn=ax_{n}=-a and xn=wax_{n}=w-a support KK, where 0awa10\leq a\leq w-a\leq 1 with w:=w(K)w:=w(K), i.e.

if w[0,1] then a[0,w/2]and if w[1,2] then a[w1,w/2].\text{if }w\in[0,1]\text{ then }a\in[0,w/2]\quad\text{and if }w\in[1,2]\text{ then }a\in[w-1,w/2].

Above, we take the lines containing 0 between them. This is due to Proposition 2.1, since a consequence of R(K)=1R(K)=1 with K𝔹K\subset\mathbb{B} is that 0K0\in K. In particular K𝔹{xn:axnwa}=:KK\subset\mathbb{B}\cap\{x\in\mathbb{R}^{n}:-a\leq x_{n}\leq w-a\}=:K^{\prime}, and thus V(K)V(K)V(K)\leq V(K^{\prime}).

Notice that by Fubini’s principle

V(K)=awaVn1(K{xn:xn=t})𝑑t=awaVn1(𝔹n{xn:xn=t})𝑑t=awaVn1(1t2𝔹n1)𝑑t=Vn1(𝔹n1)awa(1t2)n12𝑑t=Vn1(𝔹n1)gw(a).\begin{split}V(K^{\prime})&=\int_{-a}^{w-a}V_{n-1}(K^{\prime}\cap\{x\in\mathbb{R}^{n}:x_{n}=t\})dt\\ &=\int_{-a}^{w-a}V_{n-1}(\mathbb{B}_{n}\cap\{x\in\mathbb{R}^{n}:x_{n}=t\})dt=\int_{-a}^{w-a}V_{n-1}(\sqrt{1-t^{2}}\mathbb{B}_{n-1})dt\\ &=V_{n-1}(\mathbb{B}_{n-1})\int_{-a}^{w-a}(1-t^{2})^{\frac{n-1}{2}}dt=V_{n-1}(\mathbb{B}_{n-1})g_{w}(a).\end{split}

Since

gw(a)=(1(wa)2)n12+(1a2)n12,g_{w}^{\prime}(a)=-(1-(w-a)^{2})^{\frac{n-1}{2}}+(1-a^{2})^{\frac{n-1}{2}},

it is very easy to show that under the conditions 0awa10\leq a\leq w-a\leq 1 the function gw(a)g_{w}^{\prime}(a) is nonnegative. Thus

V(K)Vn1(𝔹n1)gw(w2)=Vn1(𝔹n1)w/2w/2(1t2)n12𝑑t=Vn1(𝔹n1)wH(1/2,(1n)/2;3/2;w2/4).\begin{split}V(K^{\prime})&\leq V_{n-1}(\mathbb{B}_{n-1})g_{w}\left(\frac{w}{2}\right)=V_{n-1}(\mathbb{B}_{n-1})\int_{-w/2}^{w/2}(1-t^{2})^{\frac{n-1}{2}}dt\\ &=V_{n-1}(\mathbb{B}_{n-1})wH(1/2,(1-n)/2;3/2;w^{2}/4).\end{split}

In order to obtain (33), we simply apply the inequality above to the set K/R(K)K/R(K) and use the nn-homogeneity of VnV_{n} and the 11-homogeneity of ww.

Evidently, equality holds for every K=𝔹n{xn:|xn|w/2}K=\mathbb{B}_{n}\cap\{x\in\mathbb{R}^{n}:|x_{n}|\leq w/2\}, for every w[0,2]w\in[0,2]. ∎

The right-hand side expression on Theorem 8.3 can be explicitly computed avoiding the use of the hypergeometric function.

Remark 8.4.

If for every nn\in\mathbb{N} we denote by

In:=(1t2)n12𝑑t,I_{n}:=\int(1-t^{2})^{\frac{n-1}{2}}dt,

it is well-known that InI_{n} can be computed recursively in terms of In2I_{n-2} when integrating by parts. One can derive that

I2=12(arcsint+t1t2)+c2,I3=tt33+c3,I4=18(3arcsint+t1t2(52t2))+c4,\begin{split}&I_{2}=\frac{1}{2}(\arcsin t+t\sqrt{1-t^{2}})+c_{2},\quad I_{3}=t-\frac{t^{3}}{3}+c_{3},\\ &I_{4}=\frac{1}{8}(3\arcsin t+t\sqrt{1-t^{2}}(5-2t^{2}))+c_{4},\end{split}

for absolute constants ci0c_{i}\geq 0, i2i\geq 2, and so on. In particular, Theorem 8.3 is equivalent to (23) when n=2n=2; if K𝒦3K\in\mathcal{K}^{3} we would then have that

V(K)π(w(K)R(K)2w(K)312);V(K)\leq\pi\left(w(K)R(K)^{2}-\frac{w(K)^{3}}{12}\right);

or if K𝒦4K\in\mathcal{K}^{4}, we would obtain that

V(K)R(K)4π3(3arcsin(w(K)2R(K))+w(K)2R(K)1w(K)24R(K)2(5w(K)22R(K)2)).\frac{V(K)}{R(K)^{4}}\leq\frac{\pi}{3}\left(3\arcsin\left(\frac{w(K)}{2R(K)}\right)+\frac{w(K)}{2R(K)}\sqrt{1-\frac{w(K)^{2}}{4R(K)^{2}}}\left(5-\frac{w(K)^{2}}{2R(K)^{2}}\right)\right).

As mentioned in Theorem (33), sets of the form K=𝔹n{xn:|xn|w/2}K=\mathbb{B}_{n}\cap\{x\in\mathbb{R}^{n}:|x_{n}|\leq w/2\} would attain equality above for every nn\in\mathbb{N} and w[0,2]w\in[0,2].

References

  • [1] P. W. Awyong, P. R. Scott, Inequalities for convex sets, J. Ineq. Pure Appl. Math. 1 (2000), no. 1, 6.
  • [2] E. Barbier, Note sur le problème de l’aiguille et le jeu du joint couvert, J. Math. Pures Appl. 5 (1860), 273–286.
  • [3] W. Blaschke, Eine Frage über konvexe Körper, Jahresber, Dtsch. Math.-Ver. 25 (1916), 121–125.
  • [4] W. Blaschke, Konvexe Bereiche gegebener konstanter Breite und kleinsten Inhalts, Math. Ann. 76 (1915), no. 4, 504–513.
  • [5] K. Böröczky Jr., M. A. Hernández Cifre and G. Salinas, Optimizing area and perimeter of convex sets for fixed circumradius and inradius, Monatsh. Math. 138 (2003), 95–110.
  • [6] R. Brandenberg, B. González Merino, A complete 3-dimensional Blaschke-Santaló-diagram, Math. Inequal. Appl. 20 (2017), no. 2, 301–348.
  • [7] R. Brandenberg, B. González Merino, Behaviour of inradius, circumradius, and diameter in generalized Minkowski spaces, RACSG 116 (2022), no. 3, 105.
  • [8] R. Brandenberg, B. González Merino, Minkowski concentricity and complete simplices, J. Math. Anal. Appl. 454 (2017), no. 2, 981–994.
  • [9] R. Brandenberg, S. König, No dimension-independent core-sets for containment under homothetics, Discrete Comput. Geom. 49 (2013), no. 1, 3–21.
  • [10] R. Brandenberg, M. Runge, Jung-type inequalities and Blaschke-Santaló diagrams for different diameter variants, in preparation.
  • [11] G. Buttazzo, A. Pratelli, An application of the continuous Steiner symmetrization to Blaschke-Santaló diagrams, ESAIM: Control, Optimisation and Calculus of Variations, 27, 36.
  • [12] A. Delyon, A. Henrot, and Y. Privat, The missing (A,D,r) diagram, Ann. de l’Institut Fourier. 72 (2022), no. 5.
  • [13] A. Erdélyi, W. Magnus, F. Oberhettinger, F. G. Tricomi, Higher transcendental functions, Vol. I. New York – Toronto – London: McGraw–Hill Book Company, Inc, 1953.
  • [14] I. Ftouhi, Complete systems of inequalities relating the perimeter, the area and the Cheeger constant of planar domains, Commun. Contemp. Math. page 2250054.
  • [15] I. Ftouhi, A. Henrot, The diagram (λ1,μ1)(\lambda_{1},\mu_{1}), preprint, hal-03311538v1, 2021.
  • [16] I. Ftouhi, J. Lamboley, Blaschke-Santaló diagram for volume, perimeter, and first Dirichlet eigenvalue, SIAM J. Math. Anal. 53 (2021), no. 2, 1670–1710.
  • [17] R. Gastaldello, A. Henrot, I. Lucardesi, About the Blaschke-Santaló diagram of area, perimeter and moment of inertia, arXiv:2307.11658.
  • [18] M. Henk, G. A. Tsintsifas, Some inequalities for planar convex figures, Elem. Math. 49 (1994), no. 3, 120–125.
  • [19] M. A. Hernández Cifre, Is there a planar convex set with given width, diameter and inradius?, Amer. Math. Monthly 107 (2000), 893–900.HC2
  • [20] M. A. Hernández Cifre, Optimizing the perimeter and the area of convex sets with fixed diameter and circumradius, Arch. Math. 79 (2002), 147–157.
  • [21] M. A. Hernández Cifre, J. A. Pastor, G. Salinas Martínez, S. Segura Gomis, Complete systems of inequalities for centrally symmetric convex sets in the n-dimensional space, Arch. Inequal. Appl. 1 (2003), no. 2, 155–167.
  • [22] M. A. Hernández Cifre, G. Salinas, S. Segura Gomis, Complete systems of inequalities, J. Ineq. Pure Appl. Math. 2 (2001), no. 1-10, 1–12.
  • [23] M. A. Hernández Cifre, S. Segura Gomis, The missing boundaries of the Santaló diagrams for the cases (d,w,R)(d,w,R) and (w,R,r)(w,R,r), Discr. Comp. Geom. 23 (2000), 381–388.
  • [24] H. Jung, Über die kleinste Kugel, die eine räumliche Figur einschlieβ\betat, J. Reine Angew. Math. 123 (1901), 241–257.
  • [25] T. Kubota, Einige Ungleischheitsbezichungen über Eilinien und Eiflächen, Sci. Rep. of the To~\tilde{o}hoku Univ. Ser. (1), 12 (1923), 45–65.
  • [26] T. Kubota, Eine Ungleischheit für Eilinien, Math. Z. 20 (1924), 264–-266.
  • [27] H. Lebesgue, Sur le problème des isopérimètres et sur les domaines de largeur constante, Bulletin de la Société Mathématique de France, 7: 72–76.
  • [28] I. Lucardesi, D. Zucco, On Blaschke–Santaló diagrams for the torsional rigidity and the first Dirichlet eigenvalue, Ann. di Mat. Pura ed Appl. (1923-) 201 (2022), no. 1, 175–201.
  • [29] Y. Martínez-Maure, De nouvelles inégalités géométriques pour les hérissons, Arch. Math. 72 (1999), 444–453.
  • [30] H. Martini, L. Montejano, D. Oliveros, Bodies of constant width, Springer International Publishing, 2019.
  • [31] R. Osserman, The isoperimetric inequality, Bull. Amer. Math. Soc. 84 (1978), no. 6, 1182–1238.
  • [32] J. Pál, Ein Minimal problem für Ovale, Math. Ann. 83 (1921), 311–319.
  • [33] J. R. Sangwine-Yager, The missing boundary of the Blaschke diagram, Amer. Math. Monthly 96 (1989), 233–237.
  • [34] L. Santaló, Sobre los sistemas completos de desigualdades entre tres elementos de una figura convexa plana, Math. Notae 17 (1961), 82–104.
  • [35] R. Schneider, Convex bodies: the Brunn-Minkowski theory. Second edition. Cambridge University Press, Cambridge 2014.
  • [36] P. R. Scott, A family of inequalities for convex sets, Bull. Austral. Math. Soc. 20 (1979), 237–245.
  • [37] P. Steinhagen, Über die gröβ\betate Kugel in einer konvexen Punktmenge, Abh. Hamb. Sem. Hamburg 1 (1921), 15–26.
  • [38] L. Ting, J. B. Keller, Extremal convex planar sets, Discrete Comput. Geom. 33 (2005), no. 3, 369–393.