# Extremizers and stability of the Betke–Weil inequality

Ferenc A. Bartha111University of Szeged, Dugonics ter 13, H-6720 Szeged,
Hungary. E-mail<EMAIL_ADDRESS>Ferenc Bencs222Alfréd Rényi
Institute of Mathematics, Reáltanoda u. 13-15, H-1053 Budapest, Hungary.
E-mail<EMAIL_ADDRESS>Károly J. Böröczky333Alfréd Rényi Institute of
Mathematics, Reáltanoda u. 13-15, H-1053 Budapest, Hungary.
Central European University, Nador utca 9, H-1051 Budapest, Hungary.
E-mail<EMAIL_ADDRESS>Daniel Hug444Karlsruhe Institute of
Technology (KIT), D-76128 Karlsruhe, Germany. E-mail<EMAIL_ADDRESS>
(To the memory of Ulrich Betke and Wolfgang Weil)

###### Abstract

Let $K$ be a compact convex domain in the Euclidean plane. The mixed area
$A(K,-K)$ of $K$ and $-K$ can be bounded from above by
$1/(6\sqrt{3})L(K)^{2}$, where $L(K)$ is the perimeter of $K$. This was proved
by Ulrich Betke and Wolfgang Weil (1991). They also showed that if $K$ is a
polygon, then equality holds if and only if $K$ is a regular triangle. We
prove that among all convex domains, equality holds only in this case, as
conjectured by Betke and Weil. This is achieved by establishing a stronger
stability result for the geometric inequality $6\sqrt{3}A(K,-K)\leq L(K)^{2}$.
Keywords. Geometric inequality, Brunn–Minkowski theory, Minkowski inequality,
perimeter, mixed area, stability result
MSC. Primary 52A39, 52A40, 52A10; Secondary 52A25, 52A38.

##  1 Introduction

For convex domains $K,M$ (compact convex sets with non-empty interior) in
${\mathbb{R}}^{2}$, let $L(K)$ be the perimeter of $K$, let $A(K)$ be the area
of $K$, and let $A(K,M)$ denote the mixed area of $K$ and $M$ (see R.
Schneider [16] or Section 2). In [5], U. Betke and W. Weil proved the
following theorem.

######  Theorem 1.1 (Betke, Weil (1991)).

If $K,M\subset{\mathbb{R}}^{2}$ are convex domains, then

$L(K)L(M)\geq 8\,A(K,M)$ (1.1)

with equality if and only if $K$ and $M$ are orthogonal (possibly degenerate)
segments.

This result has been generalized to higher dimensions in [6], where also
various improvements in the sense of stability results have been obtained.
What makes the variational analysis of (1.1) convenient is the fact that $K$
and $M$ can be varied independently of each other and the dependence on $K$
and $M$ is Minkowski linear (in the Euclidean plane). In [5], U. Betke and W.
Weil also considered the case where $M=-K$ and found the following sharp
geometric inequality.

######  Theorem 1.2 (Betke, Weil (1991)).

If $K$ is a convex domain in ${\mathbb{R}}^{2}$, then

$L(K)^{2}\geq 6\sqrt{3}\,A(K,-K).$ (1.2)

In addition, if $K$ is a polygon, then equality holds if and only if $K$ is a
regular triangle.

It is clear from the continuity of the involved functionals that it is
sufficient to establish this inequality for convex polygons to deduce it for
general convex domains in the plane. However, it has remained an open problem
to characterize the equality case in (1.2) among all convex domains. We
resolve this problem by proving more generally a stability version of Theorem
1.2. We refer to [6] (and in particular to the literature cited there) for a
brief introduction to stability improvements of geometric inequalities.

######  Theorem 1.3.

If $K$ is a convex domain in ${\mathbb{R}}^{2}$ and

$L(K)^{2}\leq\left(1+\varepsilon\right)6\sqrt{3}\,A(K,-K)$

for some $\varepsilon\in[0,2^{-28}]$, then there exists a regular triangle $T$
with centroid $z$ such that

$T-z\subset K-z\subset\left(1+400\sqrt{\varepsilon}\right)(T-z).$

The optimality of the stability exponent $\frac{1}{2}$ of $\varepsilon$ can be
seen by considering a regular triangle $T$ of edge length 2. Then we add over
each edge $E_{i}$ of $T$ an isosceles triangle with height
$\sqrt{\varepsilon}$ which has the side $E_{i}$ in common with $T$ (for
$i=1,2,3$). For the resulting hexagon $H$ we have
$L(H)^{2}-6\sqrt{3}A(H,-H)=36\varepsilon$. However, if $d_{\rm tr}(H)$ is the
minimal number $\rho\geq 0$ for which there is a regular triangle $T_{0}$ with
centroid $z$ such that $T_{0}-z\subset H-z\subset(1+\rho)(T_{0}-z)$, then it
is easy to check that $d_{\rm tr}(H)\geq\sqrt{\varepsilon}$.

######  Corollary 1.4.

Equality holds in (1.2) if and only if $K$ is a regular triangle.

In [5], Betke and Weil also discuss an application of their Theorem 1.2 to an
inequality for characteristics of a planar Boolean model. As a consequence of
Corollary 1.4, the equality condition for the lower bound provided in [5,
Theorem 3] now turns into an ‘if and only if’ statement.

For the proof of the inequality (1.2), it is sufficient to consider (convex)
polygons with at most $k$ vertices, for any fixed $k\geq 3$. This task was
accomplished by Betke and Weil and we add the observation (extracted from an
adaptation of their argument) that for a polygon $P$ which is not a regular
$k$-gon with an odd number of sides, there exist polygons $P^{\prime}$ with at
most $k$ vertices arbitrarily close to $P$ such that

$\frac{L(P^{\prime})^{2}}{A(P^{\prime},-P^{\prime})}<\frac{L(P)^{2}}{A(P,-P)},$

see Proposition 5.1. While this can be used to prove the inequality, it does
not give control over the equality cases. To determine all extremal sets, we
show that if $K$ is a convex domain which is not too far from a regular
triangle, then the inequality (1.2) can be strengthened to a stability result,
that is, we show that if $K$ also satisfies

$L(K)^{2}\leq(1+\varepsilon)6\sqrt{3}A(K,-K),$

then $K$ is $\varepsilon$ close to a regular triangle (if $\varepsilon>0$ is
small enough). This local stability result is stated and proved in Section 4
(see Proposition 4.1). An outline of the proof of Propositon 4.1, which is
divided into six steps, is given at the beginning of the proof. A major
geometric idea, underlying the argument is to approximate $K$ from inside by a
triangle $T\subset K$ with maximal area. With $T$ and $K$ we associate
hexagons $H_{1},H_{2}$ with $H_{2}\supset K$ and $T\subset H_{1}\subset K$.
Another hexagon $H_{0}$ is derived from $H_{1}$ so that the perimeter is
minimized. Then we show that

$L(K)^{2}-6\sqrt{3}A(K,-K)\geq L(H_{0})^{2}-6\sqrt{3}A(H_{2},-H_{2})\geq 0.$

The fact that the right side is nonnegative is far from obvious. More
generally, we use a variational argument and validated numerics to establish a
lower bound which involves five parameters which determine the shapes of $T$
and $H_{2}$ (see Lemma 3.1). In the course of the proofs, we have to determine
various mixed areas of polygons. These mixed areas are obtained by a classical
formula due to Minkowski and by a more recent one which is due to Betke [4]
and was first applied in [5].

##  2 Notation and mixed area

For $p_{1},\ldots,p_{\ell}\in{\mathbb{R}}^{2}$, $\ell\in{\mathbb{N}}$, we
write $[p_{1},\ldots,p_{\ell}]$ to denote the convex hull of the point set
$\\{p_{1},\ldots,p_{\ell}\\}$. In particular, $[p_{1},p_{2}]$ is the segment
connecting $p_{1}$ and $p_{2}$, and if $p_{1},p_{2},p_{3}$ are not collinear,
then $[p_{1},p_{2},p_{3}]$ is the triangle with vertices $p_{1},p_{2},p_{3}$.
In addition, the positive hull of $p_{1},p_{2}\in{\mathbb{R}}^{2}$ is given by
${\rm
pos}\\{p_{1},p_{2}\\}=\\{\alpha_{1}p_{1}+\alpha_{2}p_{2}:\alpha_{1},\alpha_{2}\geq
0\\}$. The scalar product of $x,y\in{\mathbb{R}}^{2}$ is denoted by $\langle
x,y\rangle$, and the corresponding Euclidean norm of $x$ is $\|x\|=\langle
x,x\rangle^{1/2}$. In addition, the determinant of the $2\times 2$ matrix with
columns $x,y\in{\mathbb{R}}^{2}$ is denoted by $\det(x,y)$. The space of
compact convex sets in ${\mathbb{R}}^{2}$ is equipped with the Hausdorff
metric.

In the following, by a polygon we always mean a convex set. For a (convex)
polygon $P$ in ${\mathbb{R}}^{2}$, let $\mathcal{U}(P)$ denote the finite set
of exterior unit normals to the sides of $P$. For $u\in\mathcal{U}(P)$, we
write $S_{P}(u)$ to denote the length of the side of $P$ with exterior normal
$u$. As usual, the support function $h_{K}$ of a compact convex set
$K\subset{\mathbb{R}}^{2}$ is defined by $h_{K}(x)=h(K,x)=\max\\{\langle
x,y\rangle:y\in K\\}$ for $x\in{\mathbb{R}}^{2}$.

We recall and will use repeatedly two formulas which allow us to calculate and
analyze the mixed area $A(P,Q)$ of two polygons $P$ and $Q$. The first is due
to Minkowski (see [16, 12]) and states that

$A(P,Q)=\frac{1}{2}\sum_{u\in\mathcal{U}(P)}h_{Q}(u)S_{P}(u).$ (2.1)

Since $h_{-P}(u)=h_{P}(-u)$ for $u\in{\mathbb{R}}^{2}$, (2.1) implies that

$A(P,-P)=\frac{1}{2}\sum_{u\in\mathcal{U}(P)}h_{P}(-u)S_{P}(u).$ (2.2)

Since $A(P,-P)=A(-P,P)$ (see also below) and $S_{-P}(u)=S_{P}(-u)$, we also
have

$A(P,-P)=\frac{1}{2}\sum_{u\in\mathcal{U}(P)}h_{P}(u)S_{P}(-u).$

For instance, if $P$ is a triangle it follows from (2.2) that $A(P,-P)=2A(P)$.

Another useful formula was established much later by Betke [4]. If $w$ is a
unit vector with $w\not\in\mathcal{U}(P)\cup\mathcal{U}(-Q)$, then

$2\cdot
A(P,Q)=\sum_{\begin{subarray}{c}u\in\mathcal{U}(P),v\in\mathcal{U}(Q)\\\
w\in{\rm pos}\\{u,-v\\}\end{subarray}}|\det(u,v)|S_{P}(u)S_{Q}(v).$ (2.3)

In particular, (2.3) yields that if $w\not\in\mathcal{U}(P)$ for a (fixed)
unit vector $w$, then

$A(P,-P)=\sum_{\begin{subarray}{c}\\{u,v\\}\in\mathcal{U}(P)\\\ w\in{\rm
pos}\\{u,v\\}\end{subarray}}|\det(u,v)|S_{P}(u)S_{P}(v),$ (2.4)

where the summation extends over all subsets of $\mathcal{U}(P)$ of
cardinality two for which $w\in{\rm pos}\\{u,v\\}$ holds (the factor $2$ from
the preceding formula cancels, since we do not consider ordered pairs). It is
formula (2.4) that was used in a clever way by Betke and Weil [5] to prove the
Betke–Weil inequality stated in Theorem 1.2.

Minkowski proved that the notion of mixed area can be extended to any pair
$K,M$ of compact convex sets in ${\mathbb{R}}^{2}$ (see [16, 12]), and for
compact convex sets $K_{1},K_{2},M\subset{\mathbb{R}}^{2}$ and
$\alpha_{1},\alpha_{2}\geq 0$, it is known that

$\displaystyle A(K,M)$ $\displaystyle=A(M,K),$ $\displaystyle
A(K+z_{1},M+z_{2})$ $\displaystyle=A(K,M)\mbox{ \ for
$z_{1},z_{2}\in{\mathbb{R}}^{2}$,}$ $\displaystyle A(\Phi K,\Phi M)$
$\displaystyle=|\det\Phi|\cdot A(K,M)\mbox{ \ for $\Phi\in{\rm
GL}(2,{\mathbb{R}})$,}$ $\displaystyle A(K,K)$ $\displaystyle=A(K),\mbox{ \
which is the area of }K,$ $\displaystyle A(\alpha_{1}K_{1}+\alpha_{2}K_{2},M)$
$\displaystyle=\alpha_{1}A(K_{1},M)+\alpha_{2}A(K_{2},M),$ $\displaystyle
A(K_{1},M)$ $\displaystyle\leq A(K_{2},M)\mbox{ \ if }K_{1}\subset K_{2}.$

We note that it is a subtle issue to decide under which conditions on convex
polygons $P\subset Q$ the inequality $A(P,-P)\leq A(Q,-Q)$ is strict. For
example, let $P$ be a triangle with its centroid at the origin $o$. Let
$v_{1},v_{2},v_{3}$ denote the vertices of $P$, and let $Q$ be the hexagon
with vertices $v_{1},v_{2},v_{3},-v_{1},-v_{2},-v_{3}$. Then $P\subset Q$ and
$P\neq Q$, actually $A(Q)=2A(P)$, and still we have
$A(P,-P)=2A(P)=A(Q)=A(Q,-Q)$.

Finally, we recall that $A(\cdot,\cdot)$ is additive (a valuation) in both
arguments. By this we mean that if $K,M,L$ are compact convex sets in the
plane and $K\cup M$ is also convex, then

$A(K\cup M,L)+A(K\cap M,L)=A(K,L)+A(M,L).$

By symmetry the same property holds for the second argument.

##  3 An auxiliary result for associated hexagons

Let $T\subset K$ denote a triangle of maximal area contained in $K$. Let
$v_{1},v_{2},v_{3}$ be the vertices of $T$, let $a_{i}$ be the side opposite
to $v_{i}$, whose length is also denoted by $a_{i}$ for $i=1,2,3$, and let
$h_{i}$ be the height of $T$ corresponding to $a_{i}$. Then we have

$2A(T)=a_{1}h_{1}=a_{2}h_{2}=a_{3}h_{3}.$

3We observe that the line passing through $v_{i}$ and parallel to the side
$a_{i}$ is a supporting line to $K$, by the maximality of the area of $T$. The
width of $K$ orthogonal to $a_{i}$ can be expressed in the form
$(1+t_{i})h_{i}$ for some $t_{i}\in[0,1]$, where $t_{i}\leq 1$ follows since
$T$ has maximal area among all triangles in $K$. (The width of $K$ orthogonal
to $a_{i}$ equals the length of the projection of $K$ to a line orthogonal to
$a_{i}$.) It follows that $K$ is contained in a circumscribed hexagon $H_{2}$
such that, for $i=1,2,3$, $H_{2}$ has two sides parallel to $a_{i}$, one of
which contains $v_{i}$ and has length $(t_{j}+t_{k})a_{i}$,
$\\{i,j,k\\}=\\{1,2,3\\}$, and the opposite side has length $(1-t_{i})a_{i}$
(see Figure 3.1). These assertions follow by elementary geometry from the
similarity of corresponding triangles. In fact, with the notation from Figure
3.1 we have

${\|w_{32}^{\prime}-w_{32}\|}=\frac{a_{2}}{a_{3}}\cdot{\|w_{32}-v_{3}\|}\quad\text{and}\quad{\|w_{32}^{\prime}-w_{32}\|}=\frac{a_{2}}{h_{2}}\cdot{t_{2}h_{2}},$

hence

$\|w_{32}^{\prime}-w_{32}\|=t_{2}a_{2},\quad\|w_{32}-v_{3}\|=t_{2}a_{3},$

and similarly for permutations of the indices. Moreover,

$\frac{\|w_{32}^{\prime}-w_{12}^{\prime}\|}{a_{2}}=\frac{h_{2}(1+t_{2})}{h_{2}},$

so that $\|w_{32}^{\prime}-w_{12}^{\prime}\|=(1+t_{2})a_{2}$, and hence

$\|w_{32}-w_{12}\|=(1+t_{2})a_{2}-\|w_{32}^{\prime}-w_{32}\|-\|w_{12}^{\prime}-w_{12}\|=(1-t_{2})a_{2}.$

Next, for $i=1,2,3$ and $\\{i,j,k\\}=\\{1,2,3\\}$ we choose a point
$q_{i}\in[w_{ji},w_{ki}]\cap K$. Then we define the (possibly degenerate)
hexagon $H_{1}=[v_{1},q_{2},v_{3},q_{1},v_{2},q_{3}]\subset K$. In addition,
let $p_{i}$ be the point on the line determined by $[w_{ji},w_{ki}]$ (and
parallel to $a_{i}$) which lies on the perpendicular bisector of the side
$a_{i}$ of $T$, and let $H_{0}$ be the hexagon with vertices
$v_{1},p_{2},v_{3},p_{1},v_{2},p_{3}$ (see Figure 3.1). Note that in general
$H_{0}$ may not be convex, but the restricted choice of parameters encountered
in the following will always ensure that $H_{0}$ is convex.

$a_{1}$

$a_{2}$

$a_{3}$

$q_{1}$

$p_{1}$

$(1-t_{1})a_{1}$

$q_{2}$

$p_{2}$

$(1-t_{2})a_{2}$

$q_{3}$

$p_{3}$

$(1-t_{3})a_{3}$

$h_{1}$

$h_{2}$

$h_{3}$

$t_{1}a_{2}$$t_{3}a_{2}$

$v_{2}$

$w_{21}$

$w_{23}$

$v_{1}$

$w_{13}$

$w_{12}$

$w_{12}^{\prime}$

$t_{3}a_{1}$$t_{2}a_{1}$

$v_{3}$

$w_{32}$

$w_{32}^{\prime}$

$w_{31}$

$t_{1}a_{3}$$t_{2}a_{3}$

Figure 3.1: Illustration of the geometric construction with the triangle
$T=[v_{1},v_{2},v_{3}]$, the outer hexagon
$H_{2}=[w_{32},w_{31},w_{21},w_{23},w_{13},w_{12}]$, the inscribed hexagon
$H_{1}=[v_{1},q_{2},v_{3},q_{1},v_{2},q_{3}]$ and the hexagon
$H_{0}=[v_{1},p_{2},v_{3},p_{1},v_{2},p_{3}]$. In particular, we have
$\|w_{ij}-v_{i}\|=t_{j}a_{i}$ and $\|w_{ij}-w_{kj}\|=(1-t_{j})a_{j}$ for
$\\{i,j,k\\}=\\{1,2,3\\}$.

As $H_{1}\subset K\subset H_{2}$ and $L(H_{0})\leq L(H_{1})$, we have

$\displaystyle{L(K)^{2}}-6\sqrt{3}\,{A(K,-K)}$ (3.1)
$\displaystyle\qquad\qquad\geq{L(H_{1})^{2}}-6\sqrt{3}\,{A(H_{2},-H_{2})}\geq{L(H_{0})^{2}}-6\sqrt{3}\,{A(H_{2},-H_{2})}.$

Clearly, $T$ is also a triangle of maximal area contained in $H_{1}$. As among
convex domains of given area, the maximal area of an inscribed triangle is the
smallest for ellipses (see Blaschke [3], Sas [15] and Schneider [16, Theorem
10.3.3]), we have

$A(T)\geq\frac{3\sqrt{3}}{4\pi}\cdot A(H_{1})>0.4\cdot
A(H_{1})=0.4(1+t_{1}+t_{2}+t_{3})A(T),$

and hence

$t_{1}+t_{2}+t_{3}<1.5.$ (3.2)

The following lemma is the basis for obtaining better bounds on
$t_{1},t_{2},t_{3}$ if we know that ${L(K)^{2}}-6\sqrt{3}\,{A(K,-K)}$ is
small.

######  Lemma 3.1.

If $a_{1}=2$, $a_{2},a_{3}\in[2,2+\frac{1}{6}]$ and
$t_{1},t_{2},t_{3}\in[0,\frac{1}{6}]$, then the hexagons $H_{0}$ and $H_{2}$
constructed as above satisfy

$L(H_{0})^{2}-6\sqrt{3}\,A(H_{2},-H_{2})\geq(a_{2}-2)^{2}+(a_{3}-2)^{2}+(t_{1}-t_{0})^{2}+(t_{2}-t_{0})^{2}+(t_{3}-t_{0})^{2}$

for $t_{0}=(t_{1}+t_{2}+t_{3})/3$.

Remark In the lemma, we do not need $K$, only the triangle $T$ and
$t_{1},t_{2},t_{3}\geq 0$ are required to define $H_{0}$ and $H_{2}$.
Moreover, although $H_{0}$ will be convex in the situation of the lemma, this
will not be needed in the argument.

###### Proof.

By the translation invariance of the mixed area, we can assume that $v_{2}$ is
the origin. Then we obtain from (2.2) that

$\displaystyle A(H_{2},-H_{2})$
$\displaystyle=\frac{1}{2}\left\\{(1-t_{2})a_{2}\cdot
0+(t_{1}+t_{2})a_{3}t_{3}h_{3}+(1-t_{1})a_{1}h_{1}\right.$
$\displaystyle\left.\qquad\qquad+(t_{1}+t_{3})a_{2}h_{2}(1+t_{2})+(1-t_{3})a_{3}h_{3}+(t_{2}+t_{3})a_{1}h_{1}t_{1}\right\\}$
$\displaystyle=2A(T)(1+t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1}).$ By Heron’s formula,
$\displaystyle 2A(T)$ $\displaystyle=a_{1}h_{1}=a_{2}h_{2}=a_{3}h_{3}$
$\displaystyle=\frac{1}{2}\sqrt{(a_{1}+a_{2}+a_{3})(-a_{1}+a_{2}+a_{3})(a_{1}-a_{2}+a_{3})(a_{1}+a_{2}-a_{3})},$
and in addition we have $\displaystyle L(H_{0})$
$\displaystyle=\sum_{i=1}^{3}\sqrt{a_{i}^{2}+4t_{i}^{2}h_{i}^{2}}.$

Setting $b_{i}=2t_{i}h_{i}=\frac{4\,A(T)t_{i}}{a_{i}}$ for $i=1,2,3$, it
follows from the Minkowski inequality (or equivalently, the triangle
inequality for $(a_{i},b_{i})$, $i=1,2,3$) that

$\displaystyle L(H_{0})^{2}$
$\displaystyle=\left(\sum_{i=1}^{3}\sqrt{a_{i}^{2}+b_{i}^{2}}\right)^{2}$
$\displaystyle\geq(a_{1}+a_{2}+a_{3})^{2}+(b_{1}+b_{2}+b_{3})^{2}$
$\displaystyle=(a_{1}+a_{2}+a_{3})^{2}+16\,A(T)^{2}(t_{1}/a_{1}+t_{2}/a_{2}+t_{3}/a_{3})^{2}$
$\displaystyle=:f_{1}(a_{2},a_{3},t_{1},t_{2},t_{3}).$

For the subsequent analysis, we set
$f_{2}(a_{2},a_{3},t_{1},t_{2},t_{3}):=16A(T)^{2}$, hence

$f_{2}(a_{2},a_{3},t_{1},t_{2},t_{3})=(a_{1}+a_{2}+a_{3})(-a_{1}+a_{2}+a_{3})(a_{1}-a_{2}+a_{3})(a_{1}+a_{2}-a_{3})$

and

$f_{1}(a_{2},a_{3},t_{1},t_{2},t_{3})=(a_{1}+a_{2}+a_{3})^{2}+f_{2}(a_{2},a_{3},t_{1},t_{2},t_{3})(t_{1}/a_{1}+t_{2}/a_{2}+t_{3}/a_{3})^{2},$

and finally we set

$f(a_{2},a_{3},t_{1},t_{2},t_{3}):=f_{1}(a_{2},a_{3},t_{1},t_{2},t_{3})-3\sqrt{3}\sqrt{f_{2}(a_{2},a_{3},t_{1},t_{2},t_{3})}\,(1+t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1}).$

Thus we obtain

$L(H_{0})^{2}-6\sqrt{3}\cdot A(H_{2},-H_{2})\geq
f(a_{2},a_{3},t_{1},t_{2},t_{3}),$ (3.3)

In the following, we consider

$\displaystyle W$
$\displaystyle:=\left\\{(a_{2},a_{3},t_{1},t_{2},t_{3})^{\top}\in{\mathbb{R}}^{5}:\,\mbox{$a_{2},a_{3}\in[2,2+\frac{1}{6}]$
and $t_{1},t_{2},t_{3}\in[0,\frac{1}{6}]$}\right\\},$ $\displaystyle z_{t}$
$\displaystyle:=(2,2,t,t,t)^{\top},\quad t\in[0,\tfrac{1}{6}],$

and the orthonormal basis

$\displaystyle e_{1}$ $\displaystyle=(1,0,0,0,0)^{\top},\quad
e_{2}=(0,1,0,0,0)^{\top},\quad
e_{3}=\left(0,0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0\right)^{\top},$
$\displaystyle e_{4}$
$\displaystyle=\left(0,0,\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\right)^{\top},\quad
e_{5}=\left(0,0,\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)^{\top}.$

We write $Df$ for the derivative and $D^{2}f$ to denote the Hessian of $f$.
Using a computer algebra system (for convenience) or direct calculations, we
obtain that if $t\in[0,\tfrac{1}{6}]$, then

$f(z_{t})=0,\quad Df(z_{t})=o.$ (3.4)

By the Taylor formula and (3.4) there is a $\xi\in(0,1)$ such that

$f(x)=\frac{1}{2}\,\left\langle
x-z_{t},D^{2}f(z_{t}+\xi(x-z_{t}))(x-z_{t})\right\rangle,\quad x\in W.$

By relation (3.3), Lemma 3.1 follows once we have shown that

$f(x)\geq\|x-z_{t}\|^{2}$ (3.5)

for $x=(a_{2},a_{3},t_{1},t_{2},t_{3})^{\top}\in W$ and
$t=(t_{1}+t_{2}+t_{3})/3$.

Since for $x=(a_{2},a_{3},t_{1},t_{2},t_{3})^{\top}\in W$, we have $x-z_{t}\in
e_{5}^{\perp}$ (we write $e_{5}^{\perp}$ for the orthogonal complement of
$e_{5}$) with $t=(t_{1}+t_{2}+t_{3})/3$ and $z_{t}+\xi(x-z_{t})\in W$, the
proof will be finished if we can verify that

$\langle v,D^{2}f(\bar{x})v\rangle\geq 2\|v\|^{2},\quad\bar{x}\in W,v\in
e_{5}^{\perp}.$ (3.6)

Using a computer algebra system (such as SageMath or Maple) or tedious
calculations, we obtain that

$D^{2}f(2,2,0,0,0)=\left(\begin{array}[]{rrrrr}12&-6&0&0&0\\\ -6&12&0&0&0\\\
0&0&24&-12&-12\\\ 0&0&-12&24&-12\\\ 0&0&-12&-12&24\end{array}\right)$

has the eigenvalues $6,18,36,36,0$, and as associated pairwise orthogonal
eigenvectors, one can choose $(1,1,0,0,0)^{\top}$ to correspond to the
eigenvalue $6$, $(1,-1,0,0,0)^{\top}$ to correspond to $18$, $e_{3}$ and
$e_{4}$ to correspond to $36$, and finally $e_{5}$ to correspond to $0$.

Define the orthogonal matrix $S:=(e_{1}\ldots e_{5})\in\operatorname{O}(5)$
and write $y=(y_{1},\ldots,y_{5})^{\top}\in{\mathbb{R}}^{5}$. Further, we
define

$\tilde{f}(y_{1},\ldots,y_{5}):=f\left(\sum_{i=1}^{5}y_{i}e_{i}\right)=f(Sy),$

hence $f(x)=\tilde{f}(S^{\top}x)$, $x\in{\mathbb{R}}^{5}$. By the chain rule,
for $x\in W$ and $v\in{\mathbb{R}}^{5}$ we have

$\langle v,D^{2}f(x)v\rangle=\langle
S^{\top}v,D^{2}\tilde{f}(S^{\top}x)S^{\top}v\rangle.$

In addition, note that $\|S^{\top}v\|^{2}=\|v\|^{2}$ and

$\displaystyle S^{\top}(W)$
$\displaystyle\subset\left[2,2+\tfrac{1}{6}\right]^{2}\times\left[-\tfrac{1}{6}\sqrt{\tfrac{2}{3}},\tfrac{1}{6}\sqrt{\tfrac{2}{3}}\right]^{2}\times\left[0,\tfrac{\sqrt{3}}{6}\right]$
$\displaystyle\subset\widetilde{W}:=\left[2,2+\tfrac{1}{6}\right]^{2}\times[-0.14,0.14]^{2}\times[0,0.3].$

Here we use that for $y=S^{\top}(a_{2},a_{3},t_{1},t_{2},t_{3})^{\top}$ we
have

$\displaystyle y_{3}^{2}+y_{4}^{2}$
$\displaystyle=\frac{1}{2}(t_{1}-t_{2})^{2}+\frac{1}{6}(t_{1}+t_{2}-2t_{3})^{2}=\frac{2}{3}\frac{1}{2}\left[(t_{3}-t_{2})^{2}+(t_{3}-t_{1})^{2}+(t_{2}-t_{1})^{2}\right]$
$\displaystyle\leq\frac{2}{3}\max\\{t_{1},t_{2},t_{3}\\}^{2}=\frac{2}{3}\frac{1}{6^{2}}$

and

$0\leq
y_{5}=\frac{1}{\sqrt{3}}(t_{1}+t_{2}+t_{3})\leq\sqrt{3}\cdot\frac{1}{6}.$

Moreover, $v\in e_{5}^{\bot}$ if and only if $\langle
S^{\top}v,e_{5}^{\circ}\rangle=0$, where $e_{5}^{\circ}=(0,0,0,0,1)^{\top}$.
Hence, (3.6) follows if we can verify that

$\langle\tilde{v},D^{2}\tilde{f}(y)\tilde{v}\rangle\geq
2\|\tilde{v}\|^{2},\quad
y\in\widetilde{W},\,\tilde{v}\in(e_{5}^{\circ})^{\bot}.$

Writing $H(y):=\left(D^{2}\tilde{f}(y)_{ij}\right)_{i,j=1}^{n-1}$ for the
$(n-1)\times(n-1)$ matrix (principal minor) obtained from the $n\times n$
Hessian matrix representing $D^{2}\tilde{f}(y)$ with respect to the standard
basis $e_{1}^{\circ},\ldots,e_{5}^{\circ}$ of ${\mathbb{R}}^{5}$, we want to
verify that

$\langle\bar{v},H(y)\bar{v}\rangle\geq 2\|\bar{v}\|^{2},\quad
y\in\widetilde{W},\,\bar{v}\in{\mathbb{R}}^{4},$ (3.7)

that is to say, all eigenvalues of $H(y)$ are at least $2$. Let

$\Xi:=\left\\{(\bar{v}_{1},\ldots,\bar{v}_{4})^{\top}\in[-1,1]^{4}:\bar{v}_{i}=1\text{
for some $i\in\\{1,\ldots,4\\}$}\right\\}.$

By scaling invariance of (3.7) with respect to $\bar{v}\in{\mathbb{R}}^{4}$,
(3.7) is equivalent to

$\langle\bar{v},H(y)\bar{v}\rangle\geq 2\|\bar{v}\|^{2},\quad
y\in\widetilde{W},\,\bar{v}\in{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\Xi},$
(3.8)

Since all eigenvalues of $H((2,2,0,0,0)^{\top})$ are positive, this holds if
and only if all eigenvalues of $H(y)^{2}$ are at least $4$, for
$y\in\widetilde{W}$. The latter means that we have to show that
$\langle\bar{v},H(y)^{2}\bar{v}\rangle\geq 4\|\bar{v}\|^{2}$ for
$y\in\widetilde{W}$ and $\bar{v}\in{\mathbb{R}}^{4}$ or, equivalently,

$\|H(y)\bar{v}\|^{2}\geq 4\|\bar{v}\|^{2},\quad
y\in\widetilde{W},\,\bar{v}\in{\mathbb{R}}^{4}.$ (3.9)

Again by scaling invariance of (3.9) with respect to $\tilde{v}$, (3.9) is in
turn equivalent to

$\|H(y)\bar{v}\|^{2}\geq 4\|\bar{v}\|^{2},\quad
y\in\widetilde{W},\,\bar{v}\in\Xi.$ (3.10)

Direct rigorous numerical analysis of the eigendecomposition of the hessian
$D^{2}\tilde{f}$ may be challenging due to requiring too many subdivisions of
$\widetilde{W}$ in order to achieve the required precision [14, 11, 8, 10]. As
both, $\widetilde{W}$ and $\Xi$ are compact, finite dimensional and as the
desired inequalities, either (3.8) or (3.10), are expected to be strict, they
are well suited for being studied by rigorous numerics [13, 1, 17].

Namely, for small $\widetilde{W}^{\prime}\subset\widetilde{W}$, and
$\Xi^{\prime}\subset\Xi$, we preform the following procedure with all
computations being carried out rigorously using interval arithmetic and
automatic differentiation [13, 1, 17, 9]. First, we bound the jet of
$\tilde{f}$ up to Taylor-coefficients of degree $6$ over
$\widetilde{W}^{\prime}$. In order to increase precision and eliminate some of
the dependency issues, for a given $\widetilde{W}^{\prime}$, the degree 6 jet
of $f$ is bound both over $\widetilde{W}^{\prime}$ and over the midpoint of
$\widetilde{W}^{\prime}$. Hence, using the multivariate Taylor-expansions with
the appropriate remainder term, we obtain enhanced bounds on the Taylor-
coefficients of $\tilde{f}$ over $\widetilde{W}^{\prime}$ and, in turn, a
better enclosure of the hessian matrix $\mathrm{H}=D^{2}\tilde{f}$. Second, we
test if we can guarantee that the inequality (3.8) or (3.10) for all
$v\in\Xi^{\prime}$ holds. If that is not the case, then an adaptive bisection
scheme of $\widetilde{W}^{\prime}\times\Xi^{\prime}$ is utilized and the
arising subsets of $\widetilde{W}^{\prime}\times\Xi^{\prime}$ are processed
separately.

We have implemented our software using the package CAPD [7] and verified both
inequalities independently and successfully. The required number of subsets
(of $\widetilde{W}\times\Xi$) and the associated computational times (without
parallelization on an i7-9750) were

  * •

(3.8): $25880$ subsets, 8m14s;

  * •

(3.10): $2440$ subsets, 46s.

We note that the increased complexity of (3.8) is most likely just an artefact
of the naive computation of the inner product and could be decreased (to that
of (3.10)) by choosing a more efficient evaluation scheme. The source code and
output logs are available at [2].

In particular, both (3.8), (3.10), and, in turn, Lemma 3.1 have been verified.
∎

The following two claims will be used in the proof of Proposition 4.1.

######  Claim 3.1.

If the regular triangle $T_{0}$ of side length $b$ contains a triangle $T$
which has a side of length at most $a$, where $\frac{b}{2}\leq a\leq b$, then
$A(T)\leq\frac{a}{b}\,A(T_{0})$.

###### Proof.

Let $\ell$ be the line containing a side of $T$ of length at most $a$. We may
assume that the vertex $v$ of $T$ opposite to $\ell\cap T$ is also a vertex of
$T_{0}$.

If the distance of $\ell$ from $v$ (that is, the height of $T$) is at most
$b\sqrt{3}/2$, then we are done. Therefore we may now assume that the distance
of $\ell$ from $v$ is larger than $b\sqrt{3}/2$. Let $D_{v}$ be the circular
disc with center $v$ and radius $b\sqrt{3}/2$. Since the side $\ell\cap T$ is
disjoint from $D_{v}$, it lies in one of the two connected components of
$T_{0}\setminus D_{v}$. It follows that $T$ is contained in one of the two
triangles which are obtained by cutting $T_{0}$ into two sub-triangles by the
height emanating from $v$, thus $A(T)\leq\frac{1}{2}\,A(T_{0})$. ∎

######  Claim 3.2.

If $\varrho_{1},\varrho_{2},\varrho_{3}$ are the side lengths of a triangle
$P$ and $A(P)\geq\xi\,\varrho_{1}$, for some $\xi\in[0,\varrho_{1}]$, then

$\varrho_{2}+\varrho_{3}\geq\varrho_{1}+\frac{\xi^{2}}{\varrho_{1}}.$

###### Proof.

The height $h$ of $P$ corresponding to $\varrho_{1}$ is at least $2\xi$, and
$\varrho_{2}+\varrho_{3}$ is minimized under this condition if
$\varrho_{2}=\varrho_{3}$, thus

$\varrho_{2}+\varrho_{3}\geq
2\sqrt{\left(\frac{\varrho_{1}}{2}\right)^{2}+h^{2}}\geq\sqrt{\varrho_{1}^{2}+16\xi^{2}}=\varrho_{1}\sqrt{1+\frac{16\xi^{2}}{\varrho_{1}^{2}}}\geq\varrho_{1}+\frac{\xi^{2}}{\varrho_{1}},$

which proves the claim. ∎

##  4 Local stability

For any convex domain $K$, let $d_{\rm tr}(K)$ be the minimal $\rho\geq 0$
such that there exists a regular triangle $T$ with centroid $z$ satisfying

$T-z\subset K-z\subset(1+\rho)(T-z).$

In particular, $d_{\rm tr}(K)$ measures how close $K$ is to a suitable regular
triangle.

######  Proposition 4.1.

Suppose that $K$ is a convex domain with $d_{\rm tr}(K)\leq 6^{-2}$ and

$L(K)^{2}\leq(1+\varepsilon)6\sqrt{3}\,A(K,-K)$ (4.1)

for some $\varepsilon\in[0,(6\cdot 180)^{-2}]$. Then $d_{\rm tr}(K)\leq
400\sqrt{\varepsilon}$.

###### Proof.

Let $d_{\rm tr}(K)=\eta\leq 6^{-2}$ and $\varepsilon$ be as in the statement
of the proposition. There exists a regular triangle $T_{0}$ of side length $b$
containing $K$ such that a translate of $\frac{1}{1+\eta}\,T_{0}$ is contained
in $K$. For a triangle $T\subset K$ of maximal area contained in $K$, we have

$A(T)\geq\frac{A(T_{0})}{(1+\eta)^{2}}>\frac{A(T_{0})}{1+3\eta}.$ (4.2)

From now on, we use the notions and auxiliary constructions introduced for $K$
and $T$ at the beginning of Section 3, including the hexagons
$H_{0},H_{1},H_{2}$, the parameters $t_{1},t_{2},t_{3}\geq 0$, etc.

The main part of the proof is divided into several steps. In Step 1, we prove
that $T$ is $\sqrt{\varepsilon}$ close to a regular triangle, Step 2 shows
that $A(H_{2},-H_{2})-A(K,-K)$ is $\varepsilon$ small. Based on these
findings, Step 3 verifies that if $H_{2}$ is close to $T$ in the sense that
$\max\\{t_{1},t_{2},t_{3}\\}\leq 100\,\sqrt{\varepsilon}$ (see (4.21)), then
Proposition 4.1 holds.

The rest of the argument is indirect. Starting from Step 4, we assume that the
assumption $\max\\{t_{1},t_{2},t_{3}\\}>100\,\sqrt{\varepsilon}$ (see (4.24))
is satisfied and derive a contradiction. Under this assumption, we prove in
Step 4 that $H_{1}$ is reasonably close to $H_{0}$ in the sense that
$\|p_{i}-q_{i}\|$ is reasonably small for $i=1,2,3$ (see (4.26)). Then Step 5
verifies that $K\subset D=\frac{1}{2}\,H_{1}+\frac{1}{2}\,H_{2}$ and clearly
$D\subset H_{2}$. Finally, in Step 6, we prove that the gap between
$A(H_{2},-H_{2})$ and $A(D,-D)$ (and hence the gap between $A(H_{2},-H_{2})$
and $A(K,-K)$ by Step 5) is too large, which yields the desired contradiction.

Step 1 $T$ is $\sqrt{\varepsilon}$ close to a regular triangle.

By scaling invariance of the assertion (and symmetry), we may assume that the
side lengths of $T$ satisfy

$a_{1}\leq a_{2}\leq a_{3}\mbox{ \ and \ }a_{1}=2.$

Assuming $a_{1}<\frac{b}{2}$, we can apply Claim 3.1 with $a=b/2$ and get

$A(T)\leq\frac{1}{2}A(T_{0})\leq\frac{1}{2}(1+\eta)^{2}A(T)$

by (4.2), which is a contradiction, since $\eta\leq 6^{-2}$. Hence, we have
$\frac{b}{2}\leq a_{1}\leq b$, and another application of Claim 3.1 now yields
that

$A(T)\leq\frac{a_{1}}{b}A(T_{0})\leq\frac{a_{1}}{b}(1+3\eta)A(T),$

so that

$\frac{b}{1+3\eta}\leq a_{1}=2\leq a_{3}\leq b,$

and hence

$a_{3}\leq b\leq 2(1+3\eta)=2+6\eta\leq 2+\frac{1}{6}.$ (4.3)

It also follows from (4.2) that

$(1+t_{1}+t_{2}+t_{3})A(T)=A(H_{1})\leq A(K)\leq A(T_{0})<(1+3\eta)A(T),$

therefore

$t_{i}<\frac{1}{12}<\frac{1}{6}\mbox{ \ for $i=1,2,3$.}$ (4.4)

Thus, the condition $d_{\rm tr}(K)\leq 6^{-2}$ ensures that Lemma 3.1 can be
applied.

Hence, by a combination of (3.1) with Lemma 3.1 and the assumption (4.1) of
the proposition, we see that

$\displaystyle(a_{i}-2)^{2}$ $\displaystyle\leq
6\sqrt{3}\,A(K,-K)\varepsilon<11\,A(K,-K)\varepsilon\mbox{ \ for $i=2,3$},$
(4.5) $\displaystyle(t_{i}-t)^{2}$ $\displaystyle\leq
6\sqrt{3}\,A(K,-K)\varepsilon<11\,A(K,-K)\varepsilon\mbox{ \ for
}t=\tfrac{1}{3}(t_{1}+t_{2}+t_{3}),i=1,2,3.$ (4.6)

To estimate $A(K,-K)$, we observe that the height $h_{1}$ of $T$ corresponding
to the side $a_{1}=2$ satisfies $h_{1}\leq\sqrt{a_{3}^{2}-1}<2$, thus
$A(T)\leq\sqrt{a_{3}^{2}-1}<2$ by (4.3), and hence

$A(K,-K)\leq A(T_{0},-T_{0})=2A(T_{0})\leq 2(1+3\eta)A(T)<\frac{39}{9}<5.$
(4.7)

Now (4.5) and (4.6) imply that

$\displaystyle 2\leq a_{i}$ $\displaystyle\leq 2+7\sqrt{\varepsilon}\mbox{ \ \
\ for $i=2,3$},$ (4.8) $\displaystyle|t_{i}-t|$ $\displaystyle\leq
7\sqrt{\varepsilon}\mbox{ \ \ \ for $i=1,2,3$}.$ (4.9)

We observe that $\frac{a_{1}^{2}+a_{2}^{2}-a_{3}^{2}}{a_{2}}$ is an increasing
function of $a_{2}\geq 2$ as $a_{3}\geq a_{1}$. Writing $\alpha_{i}$ to denote
the angle of $T$ opposite to the side $a_{i}$ and using that
$7^{2}\sqrt{\varepsilon}\leq 1$, we have

$\cos\alpha_{3}=\frac{a_{1}^{2}+a_{2}^{2}-a_{3}^{2}}{2a_{1}a_{2}}\geq\frac{8-(2+7\sqrt{\varepsilon})^{2}}{8}\geq\frac{4-29\sqrt{\varepsilon}}{8}\geq\frac{1}{2}-4\sqrt{\varepsilon}>0.$

In particular, we have $\alpha_{1}\leq\alpha_{2}\leq\alpha_{3}<\pi/2$,
$\alpha_{3}\geq\pi/3$ and $\alpha_{1}\leq\pi/3$.

Since $\cos^{\prime}(s)=-\sin s\leq-\sqrt{3}/2$ if
$s\in[\frac{\pi}{3},\frac{\pi}{2}]$, the mean value theorem then implies that

$\alpha_{3}\leq\frac{\pi}{3}+\frac{2}{\sqrt{3}}\,4\sqrt{\varepsilon}\leq\frac{\pi}{3}+5\sqrt{\varepsilon}.$

Since
$\pi\leq\alpha_{1}+2\alpha_{3}\leq\alpha_{1}+2(\frac{\pi}{3}+5\sqrt{\varepsilon})$,
we conclude that

$\frac{\pi}{3}-10\sqrt{\varepsilon}\leq\alpha_{1}\leq\alpha_{2}\leq\alpha_{3}\leq\frac{\pi}{3}+5\sqrt{\varepsilon}.$
(4.10)

As a first estimate, we deduce $\tan\alpha_{3}\leq 2$, thus
$\tan^{\prime}(s)=1+(\tan s)^{2}\leq 5$ if $s\in[\frac{\pi}{3},\alpha_{3}]$.
In particular, (4.10) yields

$\sqrt{3}-50\sqrt{\varepsilon}\leq\tan\alpha_{1}\leq\tan\alpha_{2}\leq\tan\alpha_{3}\leq\sqrt{3}+25\sqrt{\varepsilon}.$
(4.11)

Now let $T^{\prime}_{1}$ be the regular triangle of edge length $2$ positioned
in such a way that the side $a_{1}$ is common with $T$ and ${\rm
int}\,T\cap{\rm int}\,T^{\prime}_{1}\neq\emptyset$. Recalling that $v_{2}$ and
$v_{3}$ are the endpoints of $a_{1}$, we write $v^{\prime}_{1}$ to denote the
third vertex of $T^{\prime}_{1}$, $z_{1}$ to denote the centroid of
$T^{\prime}_{1}$ and $m=\frac{1}{2}(v_{2}+v_{3})$ to denote the midpoint of
$a_{1}$. As $a_{3}\geq a_{2}$ and $\frac{\pi}{2}>\alpha_{3}\geq\frac{\pi}{3}$,
there exists a point $q$ such that $v^{\prime}_{1}\in[m,q]$ and
$v_{1}\in[v_{3},q]$. In addition, we consider the intersection point
$\\{p\\}=[q,m]\cap[v_{1},v_{2}]\neq\emptyset$ (see Figure 4.1).

$v_{2}$

$v_{3}$

$v_{1}^{\prime}$

$m$

$z_{1}$

$v_{1}$

$q$

$p$

$\alpha_{2}$$\alpha_{3}$

Figure 4.1: Illustration of the triangles $T=[v_{1},v_{2},v_{3}]$,
$T_{1}^{\prime}=[v_{1}^{\prime},v_{2},v_{3}]$ together with the auxiliary
points $m,p,q,z_{1}$ in the case where $p\in[v_{1}^{\prime},m]$ . In the case
$p\notin[v_{1}^{\prime},m]$ (not shown in the figure) we have
$T_{1}^{\prime}\subset T$ and $p=q=v_{1}$.

We deduce from (4.11) that

$\|q-v^{\prime}_{1}\|=\tan(\alpha_{3})-\sqrt{3}\leq 25\sqrt{\varepsilon}$

and

$\|p-m\|=\tan(\alpha_{2})\geq\tan(\alpha_{1})\geq\sqrt{3}-50\sqrt{\varepsilon},$

and if $p\in[v_{1}^{\prime},m]$, then also

$\|p-v^{\prime}_{1}\|=\|v_{1}^{\prime}-m\|-\|p-m\|=\sqrt{3}-\tan(\alpha_{2})\leq
50\sqrt{\varepsilon}.$

Therefore, $\|v^{\prime}_{1}-z_{1}\|=\frac{2}{\sqrt{3}}$ implies that in any
case

$\left(1-25\sqrt{3}\,\sqrt{\varepsilon}\right)(T^{\prime}_{1}-z_{1})\subset
T-z_{1}\subset\left(1+\frac{25\sqrt{3}}{2}\,\sqrt{\varepsilon}\right)(T^{\prime}_{1}-z_{1}).$
(4.12)

For the regular triangle

$T_{1}=z_{1}+\left(1-25\sqrt{3}\,\sqrt{\varepsilon}\right)(T^{\prime}_{1}-z_{1}),$

with centroid at $z_{1}$, we have

$T_{1}-z_{1}\subset
T-z_{1}\subset\left(1+70\sqrt{\varepsilon}\right)(T_{1}-z_{1}).$ (4.13)

In addition, let $z$ be the centroid of $T$. The argument above also shows
that

$v_{1}\in
v^{\prime}_{1}+\frac{25\sqrt{3}}{2}\,\sqrt{\varepsilon}\,(T^{\prime}_{1}-z_{1}).$

Since $z=\frac{1}{3}(v_{1}+v_{2}+v_{3})$ and
$z_{1}=\frac{1}{3}(v^{\prime}_{1}+v_{2}+v_{3})$, we get

$\begin{array}[]{rcl}z-z_{1}&\in&\frac{25\sqrt{3}}{2\cdot
3}\,\sqrt{\varepsilon}\cdot(T^{\prime}_{1}-z_{1})\subset
10\sqrt{\varepsilon}\cdot(T_{1}-z_{1}),\\\\[4.30554pt]
z_{1}-z&\in&20\sqrt{\varepsilon}\cdot(T_{1}-z_{1}),\end{array}$ (4.14)

where for the second containment we used that $-(T_{1}-z_{1})\subset
2(T_{1}-z_{1})$. In summary, (4.13) and (4.14) show that the regular triangle
$T_{1}$ is a very good approximation of $T$.

Note that $p_{i},q_{i}$ lie on the same side of $H_{2}$ (above $a_{i}$) for
$i=1,2,3$. This follows from $\alpha_{i}\leq\frac{\pi}{2}$, $a_{i}\geq 2$,
(4.3) and (4.4), which imply that
$t_{i}a_{j}\leq\frac{1}{12}(2+\frac{1}{6})<0.2<1\leq\frac{a_{k}}{2}$ for
$i,j,k\in\\{1,2,3\\}$. Moreover, for $i=1,2,3$ we have

$1.77>(1+4\sqrt{\varepsilon})(\sqrt{3}+25\sqrt{\varepsilon})\geq\frac{a_{i}}{2}\tan(\alpha_{3})\geq
h_{i}\geq\frac{a_{i}}{2}\tan(\alpha_{1})\geq\sqrt{3}-50\sqrt{\varepsilon}>1.68,$
(4.15)

and hence in particular

$2>1.77>A(T)=h_{1}>1.68>1.5.$ (4.16)

Step 2 $A(K,-K)$ and $A(H_{2},-H_{2})$ are $\varepsilon$ close, and
$L(H_{0})$, $L(H_{1})$, $L(K)$ are $\varepsilon$ close.

We deduce from (4.7) that $6\sqrt{3}\,A(K,-K)\leq 26\sqrt{3}<50$. Therefore
(4.1), Lemma 3.1, $H_{1}\subset K\subset H_{2}$ and $L(H_{0})\leq L(H_{1})$
imply that

$\displaystyle 50\varepsilon$ $\displaystyle\geq
L(K)^{2}-6\sqrt{3}\,A(K,-K)\geq L(K)^{2}-6\sqrt{3}\,A(H_{2},-H_{2})$
$\displaystyle\geq L(H_{1})^{2}-6\sqrt{3}\,A(H_{2},-H_{2})\geq
L(H_{0})^{2}-6\sqrt{3}\,A(H_{2},-H_{2})\geq 0.$ (4.17)

In particular,

$\displaystyle L(K)^{2}-6\sqrt{3}\,A(K,-K)$ $\displaystyle\leq 50\varepsilon,$
$\displaystyle 6\sqrt{3}\,A(H_{2},-H_{2})-L(K)^{2}$ $\displaystyle\leq 0,$

which yields

$A(H_{2},-H_{2})\leq A(K,-K)+5\varepsilon.$ (4.18)

Using $L(K)\geq L(H_{1})\geq L(H_{0})\geq L(T)\geq 6$ and

$\displaystyle L(K)^{2}-6\sqrt{3}\,A(H_{2},-H_{2})$ $\displaystyle\leq
50\varepsilon,$ $\displaystyle 6\sqrt{3}\,A(H_{2},-H_{2})-L(H_{1})^{2}$
$\displaystyle\leq 0,$

we get

$50\varepsilon\geq L(K)^{2}-L(H_{1})^{2}=(L(K)-L(H_{1}))(L(K)+L(H_{1}))\geq
12(L(K)-L(H_{1})).$

We deduce that

$L(K)\leq L(H_{1})+5\varepsilon.$ (4.19)

A similar argument shows that

$L(H_{1})\leq L(H_{0})+5\varepsilon.$ (4.20)

This completes Step 2.

For the remaining part of the proof, we set $\gamma=10^{2}$. In the following,
we distinguish whether

$\max\\{t_{1},t_{2},t_{3}\\}\leq\gamma\sqrt{\varepsilon}$ (4.21)

is satisfied or not. If (4.21) holds, then $H_{2}$ is $\sqrt{\varepsilon}$
close to $K$ (see the argument below).

Step 3 If (4.21) holds, then $d_{\rm tr}(K)\leq 4\gamma\sqrt{\varepsilon}$.

It follows from (4.13) and $T\subset K$ that

$T_{1}-z_{1}\subset K-z_{1}.$ (4.22)

Using (4.21) and recalling that $z$ is the centroid of $T$, we get

$K-z\subset
H_{2}-z\subset\left(1+3\max\\{t_{1},t_{2},t_{3}\\}\right)(T-z)\subset(1+3\gamma\sqrt{\varepsilon})(T-z).$

Therefore (4.13) and (4.14) imply that if (4.21) holds, then

$\displaystyle K-z_{1}$
$\displaystyle=K-z+(z-z_{1})\subset(1+3\gamma\sqrt{\varepsilon})(T-z_{1})+3\gamma\sqrt{\varepsilon}(z_{1}-z)$
$\displaystyle\subset[1+70\sqrt{\varepsilon}+3\gamma\sqrt{\varepsilon}+210\gamma\varepsilon+60\gamma\varepsilon](T_{1}-z_{1})$
$\displaystyle\subset\left(1+4\gamma\sqrt{\varepsilon}\right)(T_{1}-z_{1}).$
(4.23)

In view of (4.22), we conclude Proposition 4.1 if (4.21) holds.

It remains to consider the case where

$\max\\{t_{1},t_{2},t_{3}\\}>\gamma\sqrt{\varepsilon}.$ (4.24)

This case will finally lead to a contradiction.

It follows from (4.9) and (4.24) that

$\min\\{t_{1},t_{2},t_{3}\\}\geq 86\sqrt{\varepsilon}.$ (4.25)

Step 4 Assuming (4.24), $H_{1}$ is reasonably close to $H_{0}$.

More precisely, we claim that

$\|p_{i}-q_{i}\|<0.06<0.1\mbox{ \ \ for $i=1,2,3$},$ (4.26)

which is what we mean by saying that $H_{1}$ is “reasonably close” to $H_{0}$.

Let $\\{i,j,k\\}=\\{1,2,3\\}$, and assume that $q_{i}\neq p_{i}$ (otherwise
(4.26) readily holds). For the line $\ell_{i}$ through $p_{i}$ and $q_{i}$ and
parallel to $a_{i}$, let $\tilde{v}_{j}$ be the reflection of ${v}_{j}$
through $\ell_{i}$, and hence $p_{i}$ is the midpoint of
$[\tilde{v}_{j},v_{k}]$. For the triangle
$\widetilde{T}_{i}=[q_{i},\tilde{v}_{j},v_{k}]$, we have

$A(\widetilde{T}_{i})=2A([q_{i},p_{i},v_{k}])=\|p_{i}-q_{i}\|t_{i}h_{i}.$

Recall from (4.15) that $1.68<h_{i}<1.77$. Using (4.8) and
$t_{i}\leq\frac{1}{12}$ by (4.4), we get

$\|\tilde{v}_{j}-v_{k}\|=2\|p_{i}-v_{k}\|\leq
2\left(\frac{a_{i}}{2}+\frac{1}{12}h_{i}\right)\leq a_{i}+\frac{1}{6}h_{i}\leq
2+\frac{7}{6\cdot 180}+\frac{1}{6}\cdot 1.77\leq 2.31.$

Therefore

$A(\widetilde{T}_{i})\geq\|p_{i}-q_{i}\|t_{i}\cdot
1.68\geq\frac{\|p_{i}-q_{i}\|t_{i}}{2.31}\cdot
1.68\cdot\|\tilde{v}_{j}-v_{k}\|.$

Since

$\frac{\|p_{i}-q_{i}\|t_{i}}{2.31}\cdot 1.68\leq\frac{a_{i}}{12\cdot
2.31}\cdot 1.68<a_{i}\leq\|\tilde{v}_{j}-v_{k}\|,$

Claim 3.2 can be applied. In Combination with $t_{i}\geq 86\sqrt{\varepsilon}$
(by (4.25)), this leads to

$\displaystyle\|q_{i}-v_{j}\|+\|q_{i}-v_{k}\|$
$\displaystyle=\|q_{i}-\tilde{v}_{j}\|+\|q_{i}-v_{k}\|$
$\displaystyle\geq\|\tilde{v}_{j}-v_{k}\|+\frac{1}{2.31}\left(\frac{\|p_{i}-q_{i}\|t_{i}\cdot
1.68}{2.31}\right)^{2}$
$\displaystyle=\|p_{i}-v_{j}\|+\|p_{i}-v_{k}\|+\|p_{i}-q_{i}\|^{2}\,t_{i}^{2}\,\frac{1.68^{2}}{2.31^{3}}$
$\displaystyle\geq\|p_{i}-v_{j}\|+\|p_{i}-v_{k}\|+\|p_{i}-q_{i}\|^{2}\,\frac{86^{2}\cdot
1.68^{2}}{2.31^{3}}\,\varepsilon.$

We deduce from (4.20) that

$5\,\varepsilon\geq\|p_{i}-q_{i}\|^{2}\,\frac{86^{2}\cdot
1.68^{2}}{2.31^{3}}\,\varepsilon,$

and hence (4.26) follows.

Step 5 Assuming (4.24), we have $K\subset
D:=\frac{1}{2}\,H_{1}+\frac{1}{2}\,H_{2}$.

The polygon $D\subset H_{2}$ has twelve sides (see Figure 4.2).

$a_{1}$

$a_{2}$

$a_{3}$

$q_{1}$

$p_{1}$

$q_{2}$

$p_{2}$

$q_{3}$

$p_{3}$

$v_{2}$

$w_{21}$

$w_{23}$

$v_{1}$

$w_{13}$

$w_{12}$

$v_{3}$

$w_{32}$

$w_{31}$

Figure 4.2: Illustration for  $D$ $=\frac{1}{2}$$H_{1}$ $+\,\frac{1}{2}H_{2}$.

Six of these sides are subsets of the six sides of $H_{2}$ as
$h_{H_{1}}(u)=h_{H_{2}}(u)$ holds for any $u\in\mathcal{U}(H_{2})$, and the
other six sides of $D$ are parallel to the sides of $H_{1}$. We prove

$K\subset D$ (4.27)

indirectly, so we assume that there exists $x\in K\setminus D$, and seek a
contradiction. Then there exists $u\in\mathcal{U}(D)$ such that
$h_{K}(u)\geq\langle x,u\rangle>h_{D}(u)$. Since $x\in K\subset H_{2}$, $u$ is
normal to a side of $D$ parallel to a side of $H_{1}$. Let $u$ be normal to
$[v_{i},q_{j}]$, $i\neq j$, so that $w_{ij}$ is the vertex of $H_{2}$ where
$u$ is a normal to $H_{2}$. For

$\Delta=h_{H_{2}}(u)-h_{H_{1}}(u)=\langle u,w_{ij}-q_{j}\rangle,$

we have

$\Delta\cdot\|q_{j}-v_{i}\|=2A([q_{j},v_{i},w_{ij}])=t_{j}h_{j}\|w_{ij}-q_{j}\|.$

We have $h_{j}\geq 1.68$ by (4.15), and since $t_{i}\leq\frac{1}{12}$ by (4.4)
and $h_{i}<2$, we obtain using (4.26) that

$\|q_{j}-v_{i}\|\leq\|p_{i}-v_{i}\|+0.1\leq\frac{a_{i}}{2}+\frac{1}{12}h_{i}+0.1\leq\frac{3}{2}+\frac{1}{6}+0.1<2,$

and

$\|w_{ij}-q_{j}\|\geq\|w_{ij}-p_{j}\|-0.1\geq\frac{a_{j}}{2}-0.1\geq
1-0.1=0.9.$

Thus we get

$\Delta\geq\frac{1}{2}\,{t_{j}}\cdot 1.68\cdot 0.9\geq 60\sqrt{\varepsilon}.$

We deduce that

$\langle
u,x-q_{j}\rangle>h_{D}(u)-h_{H_{1}}(u)=\frac{1}{2}(h_{H_{2}}(u)-h_{H_{1}}(u))=\frac{\Delta}{2}\geq
30\sqrt{\varepsilon},$

therefore

$A([q_{j},v_{i},x])=\frac{\langle u,x-q_{j}\rangle\cdot\|q_{j}-v_{i}\|}{2}\geq
15\sqrt{\varepsilon}\cdot\|q_{j}-v_{i}\|.$

Note that $15\sqrt{\varepsilon}\leq 0.02$ and
$\|q_{j}-v_{i}\|\geq\|p_{j}-v_{i}\|-0.1\geq 0.9$. It follows from Claim 3.2
and $\|q_{j}-v_{i}\|\leq 2$ that

$\|x-v_{i}\|+\|x-q_{j}\|\geq\|q_{j}-v_{i}\|+\frac{1}{\|q_{j}-v_{i}\|}\cdot\left(15\sqrt{\varepsilon}\right)^{2}\geq\|q_{j}-v_{i}\|+15\varepsilon.$

Writing $\widetilde{H}$ to denote the polygon that is the convex hull of $x$
and $H_{1}$, we have that $[v_{i},x]$ and $[x,q_{j}]$ are sides of
$\widetilde{H}$, and hence

$L(K)\geq L(\widetilde{H})\geq L(H_{1})+15\varepsilon,$

contradicting (4.19). In turn, we conclude (4.27).

Step 6 $A(H_{2},-H_{2})-A(K,-K)$ is too large, assuming (4.24).

In order to calculate $A(D,-D)$ for $D=\frac{1}{2}\,H_{1}+\frac{1}{2}\,H_{2}$,
we claim that $D_{0}=\frac{1}{2}\,T+\frac{1}{2}\,H_{2}$ satisfies

$A(D,-D)=A(D_{0},-D_{0})+A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{1}t_{3}).$ (4.28)

$a_{1}$

$a_{2}$

$a_{3}$

$q_{1}$

$q_{2}$

$q_{3}$

$v_{2}$

$w_{21}$

$w_{23}$

$v_{1}$

$w_{13}$

$w_{12}$

$v_{3}$

$w_{32}$

$w_{31}$

Figure 4.3: Illustration for  $D$ $=\frac{1}{2}$$H_{1}$ $+\,\frac{1}{2}H_{2}$
and $D_{0}$$=\frac{1}{2}$$T$$+\frac{1}{2}H_{2}$.

We prove (4.28) by applying Betke’s formula (2.4) three times. Let $b_{i}$ be
the side of $D$ containing $q_{i}$ (and hence $b_{i}$ is contained in the
“long” side of $H_{2}$ parallel to $a_{i}$), and let
$\tilde{u}_{i}\in\mathcal{U}(D)$ be the normal to $D$ at the vertex $v_{i}$ of
$T$, $i=1,2,3$, and hence $-\tilde{u}_{i}$ is the exterior unit normal to
$b_{i}$. In addition, let $d_{i}$ be the diagonal of $D$ that cuts off $b_{i}$
and the two sides neighboring $b_{i}$ from $D$, and hence $d_{i}$ is parallel
to $b_{i}$, $i=1,2,3$. We write $\nu_{ij}$ to denote the exterior unit normal
to the side of $D$ cut off by and parallel to $[q_{i},v_{j}]$, $i\neq j$, and
hence $\nu_{ij}$ and $\nu_{ik}$ are the normals to the two sides neighboring
$b_{i}$, $\\{i,j,k\\}=\\{1,2,3\\}$.

Now $d_{1}$ dissects $D$ into a trapezoid and a polygon $D_{1}$ with $10$
sides, and on the way to verify (4.28), we first claim that

$A(D,-D)=A(D_{1},-D_{1})+S_{D}(\tilde{u}_{1})S_{D}(\nu_{1,2})|\det(\tilde{u}_{1},\nu_{1,2})|.$
(4.29)

To prove (4.29), we choose a unit vector $u_{1}\neq\tilde{u}_{1}$ very close
to $\tilde{u}_{1}$ and such that $\langle u_{1},v_{2}-v_{3}\rangle>0$. When we
apply Betke’s formula (2.4) to calculate the difference
$A(D,-D)-A(D_{1},-D_{1})$ using $u_{1}$ as the reference vector, we deduce
after cancellation of summands common to $A(D,-D)$ and $A(D_{1},-D_{1})$, that
the exterior unit normal $-\tilde{u}_{1}$ to the sides $b_{1}$ of $D$ and
$d_{1}$ of $D_{1}$ does not occur in either term and precisely one of the two
exterior unit normals of the two sides of $D$ neighboring $b_{1}$, in this
case $\nu_{12}$, occurs. To see this, we observe that if
$\\{i,j,k\\}=\\{1,2,3\\}$, then

$\angle(-\tilde{u}_{i},\nu_{ij})<0.16$ (4.30)

because $\angle(-\tilde{u}_{i},\nu_{ij})=\angle(q_{i}-v_{j},v_{k}-v_{j})$
satisfies $\tan\angle(-\tilde{u}_{i},\nu_{ij})\leq(t_{i}h_{i})/(1-0.06)<0.16$
using the estimates (4.26), $h_{i}<1.77$ and $t_{i}\leq\frac{1}{12}$ (cf.
(4.4)). On the one hand, if $\\{i,j\\}=\\{2,3\\}$, then (4.10), (4.30) and
$\sqrt{\varepsilon}<(6\cdot 180)^{-1}$ yield

$\angle(\tilde{u}_{1},\nu_{i1})=\angle(\tilde{u}_{1},-\tilde{u}_{i})-\angle(-\tilde{u}_{i},\nu_{i1})=\alpha_{j}-\angle(-\tilde{u}_{i},\nu_{i1})\geq\frac{\pi}{3}-10\sqrt{\varepsilon}-0.16>0.87;$

therefore, the angle of $\tilde{u}_{1}$ with any other exterior unit normal to
$D$ or $D_{1}$ is at least $0.87$. On the other hand,
$\angle(-\tilde{u}_{1},\nu_{1j})<0.16$ for $j=2,3$ by (4.30), concluding the
proof of (4.29).

Since

$S_{D}(\nu_{1,2})=\frac{1}{2}\|q_{1}-v_{2}\|,\quad
S_{D}(\tilde{u}_{1})=\frac{1}{2}(t_{2}+t_{3})a_{1},\quad|\det(\tilde{u}_{1},\nu_{12})|=\frac{t_{1}h_{1}}{\|q_{1}-v_{2}\|},$

we deduce from (4.29) that

$A(D,-D)=A(D_{1},-D_{1})+\frac{1}{2}\,t_{1}(t_{2}+t_{3})\,A(T).$ (4.31)

Next we observe that $d_{2}$ dissects $D_{1}$ into a trapezoid and a polygon
$D_{2}$ with $8$ sides. Choosing a unit vector $u_{2}\neq\tilde{u}_{2}$ very
close to $\tilde{u}_{2}$, and applying Betke’s formula (2.4) to calculate
$A(D_{1},-D_{1})-A(D_{2},-D_{2})$ with $u_{2}$ as reference vector, we
conclude as before that

$A(D_{1},-D_{1})=A(D_{2},-D_{2})+\frac{1}{2}t_{2}(t_{1}+t_{3})A(T).$

Finally, the analogous argument for a a unit vector $u_{3}\neq\tilde{u}_{3}$
very close to $\tilde{u}_{3}$ implies that

$A(D_{2},-D_{2})=A(D_{0},-D_{0})+\frac{1}{2}t_{3}(t_{1}+t_{2})A(T).$

Thus we arrive at

$A(D,-D)=A(D_{0},-D_{0})+\frac{1}{2}[t_{1}(t_{2}+t_{3})+t_{2}(t_{1}+t_{3})+t_{3}(t_{1}+t_{2})]A(T),$

which completes the proof of (4.28).

We recall that $A(T,-T)=2A(T)$,
$A(H_{2},-H_{2})=2A(T)(1+t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1})$, and observe that
$A(H_{2},-T)=A(T,-H_{2})=A(T,-T)$ by the symmetry and rigid motion invariance
of the mixed area and Minkowski’s formula (2.1). Thus (4.28) and the linearity
of the mixed area imply

$\displaystyle A(D,-D)$
$\displaystyle=A(D_{0},-D_{0})+A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{1}t_{3})$
$\displaystyle=A\left(\frac{1}{2}\,T+\frac{1}{2}\,H_{2},-\frac{1}{2}\,T-\frac{1}{2}\,H_{2}\right)+A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{1}t_{3})$
$\displaystyle=\frac{1}{4}\cdot 2A(T)+\frac{2}{4}\cdot 2A(T)+\frac{1}{4}\cdot
2A(T)(1+t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1})$
$\displaystyle\qquad\qquad+A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{1}t_{3})$
$\displaystyle=2A(T)+\frac{3}{2}A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1}).$

We deduce from (4.18), $K\subset D$ (cf. (4.27)), (4.25) and $A(T)\geq 3/2$
that

$\displaystyle 5\varepsilon$ $\displaystyle\geq A(H_{2},-H_{2})-A(K,-K)\geq
A(H_{2},-H_{2})-A(D,-D)$
$\displaystyle=2A(T)(1+t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1})-2A(T)-\frac{3}{2}A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1})$
$\displaystyle=\frac{1}{2}A(T)(t_{1}t_{2}+t_{2}t_{3}+t_{3}t_{1})\geq\frac{1}{2}\cdot\frac{3}{2}\cdot
2\cdot(86\cdot\gamma)\sqrt{\varepsilon}^{\,2}\geq\gamma^{2}\varepsilon,$

which is a contradiction, proving that (4.24) does not hold. Therefore the
argument in Step 3 proves Proposition 4.1. ∎

##  5 Proof of Theorem 1.3

Before starting the actual proof of Theorem 1.3, we recall Proposition 5.1 and
Lemma 5.2 proved in essence by Betke and Weil [5] as Lemma 1 and Lemma 2 in
[5]. We slightly modified the argument from of (2.4) and added an observation
which turns out to be useful.

######  Proposition 5.1 (Betke, Weil (1991)).

If $P$ is a polygon with $k\geq 3$ sides, and $P$ is not a regular polygon
with an odd number of sides, then there exists a polygon $P^{\prime}$ with $k$
sides and arbitrarily close to $P$ such that

$\frac{L(P^{\prime})^{2}}{A(P^{\prime},-P^{\prime})}<\frac{L(P)^{2}}{A(P,-P)}.$

###### Proof.

For a vertex $v$ of the polygon $P$, we write $N_{P}(v)$ for the normal cone
of $P$ at $v$; namely, if $u_{1},u_{2}\in\mathcal{U}(P)$ are the exterior unit
normals of the two sides meeting at $v$, then $N_{P}(v)={\rm
pos}\\{u_{1},u_{2}\\}$.

Case 1 There exist vertices $v_{1}$ and $v_{2}$ of $P$ such that
$-N_{P}(v_{1})\subset N_{P}(v_{2})$.

In this case, $P$ is not a triangle. Let $v_{3},v_{4}$ be the neighbors of
$v_{1}$. For a $v^{\prime}_{1}\in[v_{1},v_{4}]\backslash\\{v_{1},v_{4}\\}$,
let $P^{\prime}$ be obtained from $P$ by replacing the vertex $v_{1}$ by
$v^{\prime}_{1}$. In particular, there exists a unit vector $w\in-{\rm
int}\,N_{P}(v_{1})$ orthogonal to $[v^{\prime}_{1},v_{3}]$. Using this vector
$w$ in (2.4) and the property $-N_{P}(v_{1})\subset N_{P}(v_{2})$, we get
$A(P^{\prime},-P^{\prime})=A(P,-P)$, while obviously $L(P^{\prime})<L(P)$ by
strict containment.

Case 2 There exist no vertices $v_{1}$ and $v_{2}$ of $P$ such that
$-N_{P}(v_{1})\subset N_{P}(v_{2})$ as in Case 1, but there exist a vertex $v$
and a side $e$ with exterior normal $u_{0}\in\mathcal{U}(P)$ such that
$-u_{0}\in N_{P}(v)$ and $-u_{0}$ does not halve the angle of $N_{P}(v)$.

In this case, $P$ does not have any parallel sides, since we are not in Case
1. The line $\ell$ through $v$ parallel to $e$ is a support line of $P$. Let
$v_{4}$ denote the point preceding $v$ and $v_{3}$ the point following $v$ on
$\partial P$ in the clockwise order. Let $\nu_{0}$ denote the unit vector
orthogonal to $u_{0}$ such that
$(v_{4}-v)/\|v_{4}-v\|,(v_{3}-v)/\|v_{3}-v\|,\nu_{0}$ are in counter-clockwise
order on the unit circle. Let $\alpha_{4}$ denote the angle enclosed by
$v_{4}-v$ and $-\nu_{0}$ and $\alpha_{3}$ denote the angle enclosed by
$v_{3}-v$ and $\nu_{0}$. Let $u_{i}$ denote the exterior unit normal of
$[v_{i},v]$ for $i=3,4$. Since $-u_{0}$ does not halve the angle enclosed by
$u_{3},u_{4}$, we have $\alpha_{3}\neq\alpha_{4}$. We may assume that
$\alpha_{4}>\alpha_{3}$. By Fermat’s principle, moving $v$ along $\ell$ an
arbitrarily small amount in the direction of $\nu_{0}$ to $v^{\prime}$ and
denoting by $P^{\prime}$ the polygon obtained from $P$ by replacing $v$ by
$v^{\prime}$, we get $L(P^{\prime})<L(P)$. Clearly, we thus still get a convex
$k$-gon if $v^{\prime}$ is sufficiently close to $v$.

To prove $A(P,-P)=A(P^{\prime},-P^{\prime})$, we denote by $Q$ the (nonempty)
convex hull of the (common) vertices of $P$ and $P^{\prime}$ (thus
$v,v^{\prime}$ are removed). Note that in the case where $P$ (and hence also
$P^{\prime}$) is a triangle, $Q$ is a segment. We consider the triangles
$\triangle=[v_{3},v_{4},v]$ and $\triangle^{\prime}=[v_{3},v_{4},v^{\prime}]$,
hence we have $P=Q\cup\triangle$ and $P^{\prime}=Q\cup\triangle^{\prime}$. For
the segment $I=[v_{3},v_{4}]$ we have
$A(\triangle,-I)=A(\triangle,I)=A(\triangle)$. Using this and the additivity
of the mixed area in both arguments, we obtain

$A(P,-P)=A(Q,-Q)+2A(Q,-\triangle)-2A(Q,-I).$

A similar expression is obtained for $P^{\prime}$ with $\triangle$ replaced by
$\triangle^{\prime}$.

We choose $v_{5},v_{6}$ such that $e=[v_{5},v_{6}]$ and $v_{6}-v_{5}$ is a
positive multiple of $v^{\prime}-v$. If $v^{\prime}$ is sufficiently close to
$v$, then the first assumption in Claim 2 ensures that for the exterior unit
normals $u_{j}$ of the sides of $Q$ between $v_{4}$ and $v_{5}$ (in counter-
clockwise order), we have

$h_{\triangle}(u_{j})=\langle
v_{5},u_{j}\rangle=h_{\triangle^{\prime}}(u_{j})$

and that for the exterior unit normals $u_{j}$ of the sides of $Q$ between
$v_{6}$ and $v_{3}$ (in counter-clockwise order), we have

$h_{\triangle}(u_{j})=\langle
v_{6},u_{j}\rangle=h_{\triangle^{\prime}}(u_{j}).$

But then (writing $e$ also for the length of the edge $e$)

$A(Q,-\triangle)=A(Q,-\triangle^{\prime})=e\langle u_{0},-v\rangle-e\langle
u_{0},-v^{\prime}\rangle=e\langle u_{0},v^{\prime}-v\rangle=0,$

which proves the assertion.

Case 3 There exist no vertices $v_{1}$ and $v_{2}$ of $P$ such that
$-N_{P}(v_{1})\subset N_{P}(v_{2})$, and for any vertex $v$ and side $e$ with
exterior normal $u_{0}\in\mathcal{U}(P)$ such that $-u_{0}\in N_{P}(v)$ the
vector $u_{0}$ halves the angle of $N_{P}(v)$, but $P$ is not a regular
polygon with an odd number of sides.

Let $u_{1},\ldots,u_{k}$ be the exterior unit normals to the sides of $P$ in
clockwise order. We observe that no closed half plane having the origin on its
boundary contains $u_{1},\ldots,u_{k}$. We set $u_{i+pk}=u_{i}$ for any
$i=1,\ldots,k$ and $p\in{\mathbb{Z}}$.

Since there exist no vertices $v_{1}$ and $v_{2}$ of $P$ such that
$-N_{P}(v_{1})\subset N_{P}(v_{2})$, we have $u_{i}\neq-u_{j}$ for any $i,j$,
and there exists an $m\in\\{2,\ldots,k-1\\}$ such that $-u_{i}\in{\rm
int}\,{\rm pos}\\{u_{i+m-1},u_{i+m}\\}$ for any $i$. The conditions in Case 3
imply that $-u_{i}$ actually halves the angle $\angle(u_{i+m-1},u_{i+m})$
between $u_{i+m-1}$ and $u_{i+m}$.

Let us assume that $\angle(u_{m},u_{m+1})$ is minimal among the angles of the
from $\angle(u_{j},u_{j+1})$. We deduce that no $u_{j}$ lies in ${\rm
int}\,{\rm pos}\\{u_{1},-u_{m}\\}\subset{\rm pos}\\{-u_{m},-u_{m+1}\\}$;
therefore, $u_{1}=u_{2m}$. We conclude that $k=2m-1$, and hence $k$ is odd.

As $-u_{i}$ halves $\angle(u_{i+m-1},u_{i+m})$ for any $i$, and $-u_{i+m-1}$
halves $\angle(u_{i-1},u_{i})$, we deduce that
$\angle(u_{i+m-1},u_{i+m})=\angle(u_{i-1},u_{i})$ for any $i$. In turn, we
conclude that all exterior angles of $P$ are $2\pi/k$.

For any fixed $i\in{\mathbb{Z}}$, we write $e_{i}$ to denote both the side
corresponding to $u_{i}$ of $P$ and its length, and $\ell_{i}$ to denote the
line containing $e_{i}$. Since $P$ is not regular, the side lengths of $P$ are
not the same; therefore, there exists $1\leq p<q\leq k$ such that

$e_{p-1}+e_{p}\neq e_{q-1}+e_{q}.$ (5.1)

Fixing $i\in\\{1,\ldots,k\\}$, if $|t|$ is small, then let $P_{i,t}$ be the
$k$-gon bounded by the lines $\ell_{j}$,
$j\in\\{1,\ldots,k\\}\backslash\\{i\\}$ and $\ell_{i}+tu_{i}$. It follows that

$\displaystyle L(P_{i,t})$
$\displaystyle=L(P)+2t\cdot\left(\frac{1}{\sin\frac{2\pi}{k}}-\frac{1}{\tan\frac{2\pi}{k}}\right)=L(P)+\kappa\cdot
t$ and, choosing $w=-u_{i}$ in (2.4), we see that $\displaystyle
A(P_{i,t},-P_{i,t})$
$\displaystyle=A(P,-P)+(e_{i+m}+e_{i+m-1})\cdot\frac{t}{\sin\frac{2\pi}{k}}\cdot\sin\frac{\pi}{k}$
$\displaystyle=A(P,-P)+(e_{i+m}+e_{i+m-1})\cdot\varrho\cdot t.$

where $\kappa=2\frac{1-\cos\frac{2\pi}{k}}{\sin\frac{2\pi}{k}}$ and
$\varrho=\frac{\sin\frac{\pi}{k}}{\sin\frac{2\pi}{k}}$. We deduce that

$\left.\frac{d}{dt}\frac{A(P_{i,t},-P_{i,t})}{L(P_{i,t})^{2}}\right|_{t=0}=\frac{(e_{i+m}+e_{i+m-1})\cdot\varrho\cdot
L(P)-A(P,-P)\cdot 2\gamma}{L(P)^{3}}.$ (5.2)

We conclude from (5.1) and (5.2) the existence of some $i\in\\{1,\ldots,k\\}$
such that

$\left.\frac{d}{dt}\frac{A(P_{i,t},-P_{i,t})}{L(P_{i,t})^{2}}\right|_{t=0}\neq
0.$

In particular, we can choose a $t\neq 0$ with arbitrarily small absolute value
such that

$\frac{A(P_{i,t},-P_{i,t})}{L(P_{i,t})^{2}}>\frac{A(P,-P)}{L(P)^{2}},$

and hence we can choose $P^{\prime}=P_{i,t}$, completing the proof of
Proposition 5.1. ∎

For regular polygons with an odd number of sides, we have the following
estimates.

######  Lemma 5.2.

If $P$ is a regular polygon with an odd number $k\geq 5$ of sides, then

$\displaystyle d_{\rm tr}(P)$ $\displaystyle>0.25$ (5.3)
$\displaystyle\frac{L(P)^{2}}{A(P,-P)}$ $\displaystyle\geq
20\sin\frac{\pi}{5}>1.1\cdot 6\sqrt{3}.$ (5.4)

###### Proof.

We may assume that $L(P)=1$, and hence $A(P,-P)=(4k\sin\frac{\pi}{k})^{-1}$
according to (2.2), proving (5.4). For (5.3), assuming that the origin is the
centroid of $P$, we have $-P\subset\left(\cos\frac{\pi}{k}\right)^{-1}P$, and
hence $A(P,-P)\leq\left(\cos\frac{\pi}{k}\right)^{-1}A(P)$. On the other hand,
if $T_{0}\subset P$ is a regular triangle with centroid $z_{0}$ and
$P-z_{0}\subset(1+d)(T_{0}-z_{0})$, then

$A(P,-P)\geq A(T_{0},-T_{0})=2A(T_{0})\geq 2(1+d)^{-2}A(P),$

and hence $d_{\rm tr}(P)\geq\sqrt{2\cos\frac{\pi}{5}}-1>0.25$. ∎

Proof of Theorem 1.3. Let $\varepsilon\in[0,2^{-28}]$ and let
$K\subset{\mathbb{R}}^{2}$ be a convex domain with
$L(K)^{2}\leq\left(1+\varepsilon\right)6\sqrt{3}\,A(K,-K)$. Then
$0\leq\varepsilon<(6\cdot 2400)^{-2}<(6\cdot 180)^{-2}$. We distinguish two
cases.

Case 1: $d_{\rm tr}(K)\leq 6^{-2}$. Then Proposition 4.1 implies that $d_{\rm
tr}(K)\leq 400\sqrt{\varepsilon}$, and the proof is finished.

Case 2: $d_{\rm tr}(K)>6^{-2}$. We will show that in fact this case does not
occur. We fix a number $\varepsilon^{\prime}$ with
$0\leq\varepsilon<\varepsilon^{\prime}<(6\cdot 2400)^{-2}$. Since $d_{\rm
tr}(\cdot),A(\cdot,\cdot),L(\cdot)$ are continuous, there is a polygon
$P\subset{\mathbb{R}}^{2}$ with $d_{\rm tr}(P)>6^{-2}$ and
$L(P)^{2}\leq\left(1+\varepsilon^{\prime}\right)6\sqrt{3}\,A(P,-P)$. Since
$d_{\rm tr}(\cdot),A(\cdot,\cdot),L(\cdot)$ are translation invariant, $d_{\rm
tr}(\cdot)$ is scaling invariant and $K\mapsto L^{2}(K)/A(K,-K)$ (for convex
domains $K\subset{\mathbb{R}}^{2}$) is also scaling invariant, we can assume
that $o\in P$ and $A(P,-P)=1$. Let $k\geq 3$ be the number of vertices of $P$.
We write $\mathcal{P}_{k}$ for the set of all polygons
$G\subset{\mathbb{R}}^{2}$ with at most $k$ vertices, $d_{\rm tr}(G)\geq
6^{-2}$, $o\in G$, $A(G,-G)=1$ and
$L(G)^{2}\leq\left(1+\varepsilon^{\prime}\right)6\sqrt{3}$. Then in particular
we have $P\in\mathcal{P}_{k}\neq\emptyset$.

We claim that there is some $P_{0}\in\mathcal{P}_{k}$ such that
$L(P_{0})=\inf\\{L(G):G\in\mathcal{P}_{k}\\}$. For the proof, it is sufficient
to consider a minimizing sequence $P_{i}\in\mathcal{P}_{k}$ with $L(P_{i})\leq
L(P)$ for $i\in{\mathbb{N}}$. Let $B^{2}$ denote the unit disc with center at
the origin. Then $P_{i}\subset L(P)B^{2}$ for $i\in{\mathbb{N}}$. An
application of Blaschke’s selection theorem shows that the sequence $P_{i}$,
$i\in{\mathbb{N}}$, has a convergent subsequence with limit
$P_{0}\subset{\mathbb{R}}^{2}$. Since all conditions involved in the
definition of $\mathcal{P}_{k}$ are preserved under limits and $L(\cdot)$ is
continuous, we conclude that $P_{0}\in\mathcal{P}_{k}$ realizes the infimum.

Since $d_{\rm tr}(P_{0})\geq 6^{-2}>0$, $P_{0}$ is not a regular triangle.
Assuming (for the moment) that $P_{0}$ is a regular $r$-gon with an odd number
$r$, we have $k\geq r\geq 5$. Then Lemma 5.2 shows that $L(P_{0})^{2}\geq
1.1\cdot 6\sqrt{3}$. Since also
$L(P_{0})^{2}\leq(1+\varepsilon^{\prime})6\sqrt{3}$, we get
$\varepsilon^{\prime}\geq 0.1$, a contradiction.

Hence $P_{0}\in\mathcal{P}_{k}$ is a $k$-gon, but not a regular polygon with
an odd number of edges. Assume (for the moment) that $d_{\rm
tr}(P_{0})>6^{-2}$. Proposition 5.1 then shows that there is a $k$-gon $P_{1}$
such that $d_{\rm tr}(P_{1})>6^{-2}$ and

$\frac{L(P_{1})^{2}}{A(P_{1},-P_{1})}<\frac{L(P_{0})^{2}}{A(P_{0},-P_{0})}=L(P_{0})^{2}.$

Again by scaling and translation invariance, we obtain a $k$-gon $P_{2}$ for
which $d_{\rm tr}(P_{2})>6^{-2}$, $A(P_{2},-P_{2})=1$, $o\in P_{2}$ and
$L(P_{2})^{2}<L(P_{0})^{2}\leq\left(1+\varepsilon^{\prime}\right)6\sqrt{3}$,
that is, $P_{2}\in\mathcal{P}_{k}$. But this contradicts the minimality of
$L(P_{0})$. Therefore we conclude that $d_{\rm tr}(P_{0})=6^{-2}$. Since
$L(P_{0})^{2}\leq\left(1+\varepsilon^{\prime}\right)6\sqrt{3}\,A(P_{0},-P_{0})$,
it follows from another application of Proposition 4.1 that

$6^{-2}=d_{\rm tr}(P_{0})\leq 400\sqrt{\varepsilon^{\prime}}<400\cdot(6\cdot
2400)^{-1}=6^{-2},$

a contradiction. This finally shows that the present case does not occur,
which completes the argument. $\Box$

Acknowledgement. The authors are grateful for helpful discussions with Martin
Henk on the subject of the paper.

F. Bartha was supported by the grants NKFIH KKP 129877,
2020-2.1.1-ED-2020-00003, TUDFO/47138-1/2019-ITM, EFOP-3.6.2-16-2017-0015, by
the János Bolyai Research Scholarship of MTA, and by UNKP-20-5 - New national
excellence program of ITM and NKFIH.

F. Bencs is supported by the NKFIH (National Research, Development and
Innovation Office, Hungary) grant KKP-133921.

K. Böröczky was supported by research grant NKFIH K 132002.

Daniel Hug was supported by research grant HU 1874/5-1 (DFG).

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