Banach–Mazur Distance from the Parallelogram to the Affine-Regular Hexagon and Other Affine-Regular Even-Gons

We show that the Banach–Mazur distance between the parallelogram and the affine-regular hexagon is 32\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{3}{2}$$\end{document} and we conclude that the diameter of the family of centrally-symmetric planar convex bodies is just 32\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{3}{2}$$\end{document}. A proof of this fact does not seem to be published earlier. Asplund announced this without a proof in his paper proving that the Banach–Mazur distance of any planar centrally-symmetric bodies is at most 32\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{3}{2}$$\end{document}. Analogously, we deal with the Banach–Mazur distances between the parallelogram and the remaining affine-regular even-gons.


Introduction
Denote by M d the family of all centrally symmetric convex bodies of the Euclidean space E d centered at the origin o of E d . Below always when we say on a homothety, we mean that its center is at o. For any K ∈ M d and any positive λ, by λK we denote the homothety image of K with ratio λ.

Pairs of Homothetic Parallelograms Contained and Containing a Centrally Symmetric Convex Body
By an inscribed parallelogram in a convex body C we mean a parallelogram with all vertices in the boundary of C and by a circumscribed parallelogram about C we mean a parallelogram containing C whose all sides have non-empty intersections with C. By the width width(S) of a strip S between two parallel hyperplanes of E d we mean the distance of these hyperplanes. We omit an easy proof of the following lemma, whose two-dimensional version is applied in the proofs of Theorems 1 and 2.
The following proposition is applied later a few times when evaluating particular BM-distances.
Proposition. Let C ∈ M 2 . Assume that P ⊂ C ⊂ μP for a parallelogram P ∈ M 2 and a positive μ. Then there exists a parallelogram P inscribed in C such that μ P is circumscribed about C for a μ ≤ μ.
Proof. Since the thesis is obvious for μ = 1, below we assume that μ > 1.
If P in the part of P fulfills the thesis, there is nothing to prove. In the opposite case we intend to construct the required P .
Of course, there is a homothetic image P α ⊂ P of P such that for least one pair of opposite sides Z + , Z − of P α the sides μZ + , μZ − of μP α touch C from outside. Denote the other pair of sides of P α by W + , W − and assume that their order is W + , Z + , W − , Z − when we move counterclockwise.
If μW − , μW + do not touch C, then we lessen P α up to P β by moving W + , W − such that their vertices remain in Z + , Z − symmetrically closer to o up to the position F + , F − when both μF + and μF − touch C. The other two sides of the parallelogram P β are denoted by G + , G − . Clearly, G + ⊂ Z + and G − ⊂ Z − . The parallelogram μP β is circumscribed about C. Of course, when we go counterclockwise, then the order of the sides of P β is F + , G + , F − , G − .
Clearly P β is a subset of C, but it may be not inscribed in C. Take the homothetic copy P γ ⊂ C of P β such that at least one pair of opposite vertices t, v of P γ is in bd(C). Denote the other two vertices of P γ by u, w such that t, u, v, w be in this order on bd(C).
Let a, b ∈ bd(C) be the points such that w ∈ ta and w ∈ vb If u, w are not in bd(C), let us cleverly enlarge P γ up to a parallelogram P δ (not obligatory homothetic) inscribed in C such that a homothetic copy of P δ with a ratio at most μ is circumscribed about C.
Here we explain how to provide this task. For every boundary point c of bd(C) on the arc ab provide the straight lines K + 1 (c), K + 2 (c) containing cv and ct, respectively. Let K − 1 (c), K − 2 (c) be the line symmetric to K + 1 (c), K + 2 (c), respectively. Moreover, for i = 1, 2 provide the two pairs of the straight lines L + i (c) and L − i (c) supporting C and parallel to the pairs K + i (c) and If c is sufficiently close to a, then If c is sufficiently close to b, then we have the opposite inequality. Moreover, observe that when c moves on ab from a to b, then the strips L i (c) and K i (c), where i = 1, 2, change continuously. Consequently, there is at least one position c 0 of c for which width(L 1 (c 0 ))/width(K 1 (c 0 )) = width(L 2 (c 0 ))/width(K 2 (c 0 )).
Therefore the thesis of our proposition is true for the parallelograms P = Corollary. Denote the Banach-Mazur distance between C ∈ M 2 and P by μ. Assume that P ⊂ C ⊂ μP for a particular affine image P ∈ M 2 of P 4 . Then the parallelogram P is inscribed in C and μP is circumscribed about C.
The author does not know if Proposition holds true in higher dimensions for the parallelotope or the cross-polytope in place of the parallelogram. 3) is contained in P 6 and its homothetic image with ratio 3 2 contains P 6 . Consequently, δ BM (P 4 , P 6 ) ≤ 3 2 . Having in mind Proposition, in order to show that δ BM (P 4 , P 6 ) ≥ 3 2 , it is sufficient to consider only any parallelogram P = a(P 4 ) inscribed in P 6 such that a positive homothetic copy λP of P is circumscribed about P 6 , and to show that this homothety ratio λ is at least 3 2 . Denote by P the class of all such parallelograms P .

Banach-Mazur Distance
Consider a parallelogram P = pqrs from P. Some two consecutive vertices of P must be in two consecutive sides of P 6 . The reason is that in the opposite case no positive homothetic image of our P is circumscribed about P 6 in contradiction to P ∈ P. In order to fix attention, thanks to the symmetries of P 6 , we do not make our considerations narrower assuming that p ∈ v 0 v 1 and q ∈ v 1 v 2 . For the same reason, we may additionally assume that p ∈ v 0 m, where m denotes the middle of the side v 0 v 1 . Take the line y = bx passing through p. It is easy to show that p has the form ( ). An easy calculation shows that the directional coefficient of the straight line containing pq is and that the directional coefficient of the straight line containing sp . Denote by S the strip between the straight lines containing pq and rs, and by S + the narrowest strip parallel to S which contains P 6 . Denote by T the strip between the straight lines containing qr and sp, and by T + the narrowest strip parallel to T which contains P 6 .
We see that P = S ∩ T . Of course, S + ∩ T + is the parallelogram with sides parallel to the sides of P which is circumscribed about P 6 . Since we are looking only for parallelograms P ∈ P, the parallelogram S + ∩ T + should be a positive homothetic copy of P . Now, our task is to describe such parallelograms P . We find the first coordinate of the intersection of the straight line through pq with the straight liney = √ 3x. It is Next we evaluate the ratio of the first coordinate of v 1 to x 0 which is By Lemma this ratio equals to width(S + )/width(S). By the substitution of σ we get width(S + )/width(S) = √ 3 . In a similar way we obtain that width(T + )/width(T ) = Solving the equation width(S + )/width(S) = width(T + )/width(T ), we conclude that only for c = −2b √ 3b+3 its both sides are equal. We see that q ∈ nv 2 , where n is the midpoint of v 1 v 2 . Since our P is a function of b, we denote it by P (b). Substituting c = −2b √ 3b+3 into width(S + )/width(S) we obtain that the common value of w(S + )/w(S) and width(T (P (b)). This ends our task.
Every P (b) is inscribed in P 6 and every h(b)P (b) is circumscribed about P 6 (see Fig. 1). The vertices of h(b)P (b) being the images of vertices p, q, r, s of P (b) are denoted by p , q , t , u , respectively. Of course, the straight line containing the side p q of h(b)P (b) supports P 6 at v 1 .
We are considering here only the situation when the straight line containing the side s p of h(b)P (b) supports P 6 at v 5 . This holds true if and only if the directional coefficient ς of the straight line containing s p is at most the directional coefficient of the straight line containing v 5 v 0 , so when it is at most √ 3. Recall that ς = 3b+ Remark. The only two positions of P ⊂ P 6 such that P 6 ⊂ 3 2 P (besides their rotations by 60 • and 120 • and axial symmetries with respect to the lines containing v 0 v 3 , v 1 v 4 , v 2 v 5 ) are the parallelograms with vertices (1, 0), (0,

Banach-Mazur Distance Between the Parallelogram and the Affine-Regular Even-Gons
In this section we consider the BM-distances from P 4 to the regular even-gons with more than six vertices. In Theorem 2 we find these distances to P 8j and P 8j+4 , and we present the estimates from above from P 4 to P 8j+2 and P 8j+6 . Next we conjecture that the values of these two upper estimates are just the BM-distances from P 4 to P 8j+2 and P 8j+6 .
Proof. The inequalities showing that the left sides are at most the right sides result from positions of the inscribed parallelograms whose vertices are the intersections of the coordinate axes with the boundaries of P 8j , P 8j+2 , P 8j+4 , P 8j+6 , respectively. Evaluating the ratio of the smallest homothetic copy of such an inscribed parallelogram which contains our polygon from amongst P 8j , P 8j+2 , P 8j+4 , P 8j+6 , we find each value at the right sides of (I)-(IV). Let us show the opposite inequalities for (I) and (III). Ad (I). To prove that δ BM (P 4 , P 8j ) ≥ √ 2, we have to show that for any parallelogram P ∈ M 2 contained in P 8j such that P 8j ⊂ λP , where λ is positive, the inequality λ ≥ √ 2 holds true. By Proposition, in order to show that δ BM (P 4 , P 8j ) ≥ √ 2, it is sufficient to take into account only any P = a(P 4 ) inscribed in P 8j such that a positive homothetic copy λP is circumscribed about P 8j , and to show that λ ≥ √ 2. Just we may disregard the other P ⊂ P 8j . This is realized in the following Parts (α) and (β).
(α) If a parallelogram P ∈ M 2 is inscribed in P 8j and its homothetic image is circumscribed, then P is a square.