#  Torsional rigidity for tangential polygons

Grant Keady
<EMAIL_ADDRESS>
## Abstract

An inequality on torsional rigidity is established. For tangential polygons
this inequality is stronger than an inequality of Polya and Szego for convex
domains. (A survey of related work, not in the journal submission, is
presented.)

##  1 Introduction

In the most general situation $\Omega$ is a simply-connected plane domain.
However as we wish to compare our inequalities with earlier results proved for
convex domains, notably the Polya-Szego inequality (involving equation (1.3)),
we have in mind convex domains. Although Theorem 1 concerns more general
domains, our main result derived from it, Theorem 2, concerns tangential
polygons. We will denote the area of $\Omega$, $|\Omega|$, by $A$, and its
perimeter $|\partial\Omega|$ by $L$.

###  1.1 The torsion pde, Problem (P(0))

The elastic torsion problem is to find a function $u_{0}$ satisfying

$-\frac{\partial^{2}u_{0}}{\partial x^{2}}-\frac{\partial^{2}u_{0}}{\partial
y^{2}}=1\,{\rm in}\,\Omega,\,\qquad u_{0}=0\,{\rm on}\,\partial\Omega\,.$
$None$

Define the torsional rigidity of $\Omega$ as,

$Q_{0}:=\int_{\Omega}u_{0}.$ (1.1)

(This differs by a factor of 4 from definitions elsewhere, e.g in [69].) There
are many identities and inequalities concerning $Q_{0}$. For example, one
identity is

$Q_{0}=\frac{1}{4}\,\int_{\partial\Omega}(x.n)\left(\frac{\partial
u_{0}}{\partial n}\right)^{2}.$ (1.2)

Define (for convex $\Omega$), in the notation of [69],

$B_{\Omega}=\int_{\partial\Omega}\frac{1}{x\cdot{n}}.$ (1.3)

An inequality, which we will call the Polya-Szego inequality, is

$Q_{0}\geq\frac{A^{2}}{4\ B_{\Omega}}(={\rm{\ for\ tangential\ polygons\
}}\frac{A^{3}}{2L^{2}})$ (1.4)

Define

$J(v)=\int_{\Omega}\left(2v-|\nabla v|^{2}\right).$

The solution $u_{0}$ maximizes $J$ over functions vanishing on the boundary
$\partial\Omega$. (More precisely over functions
$v\in{{\mathring{W}}^{1}_{2}(\Omega)}$.)

A quantity occuring in our Theorem 1 is

$Q_{1}=\int_{\partial\Omega}\left(\frac{\partial u_{0}}{\partial
n}\right)^{2}.$ (1.5)

For tangential polygons from equation (1.2) we note that there is a very
simple equation relating $Q_{1}$ and $Q_{0}$: this is needed for our Theorem
2.

###  1.2 Problem (P($\infty$))

Problem (P($\infty$) was defined in the statement of Theorem 2.2 of [44]. Our
notation here is as in that paper. where $u_{\infty}$ solves

$-\frac{\partial^{2}u_{\infty}}{\partial
x^{2}}-\frac{\partial^{2}u_{\infty}}{\partial y^{2}}=1\,{\rm
in}\,\Omega,\,\frac{\partial u_{\infty}}{\partial
n}=-\frac{|\Omega|}{|\partial\Omega|}\,{\rm on}\,\partial\Omega\,{\rm
and}\,\int_{\partial\Omega}u_{\infty}=0.$ $None$

Define, as in [44] equations (4.9) and (4.11),

$\Sigma_{\infty}=\int_{\Omega}u_{\infty},\qquad{\rm and\ \
}\Sigma_{1}=-\int_{\partial\Omega}u_{\infty}^{2},$ (1.6)

with $u_{\infty}$ satisfying Problem (P($\infty$)).

Once again there is a variational characterisation of the solutions. This time
$u_{\infty}$ is the maximizer of $J(v)$ as one varies over functions $v$ for
which the integral of $v$ over the boundary $\partial\Omega$ is zero. The
variational approach can be used to establish the inequality

$\Sigma_{\infty}\geq Q_{0}.$ (1.7)

(See also [44] equation (4.9) for a different approach.)

The solution $u_{\infty}$ when $\Omega$ is a tangential polygon is given in
[44] equations (2.14) and (2.15).

##  2 Relations between $u_{0}$ and $u_{\infty}$

We already have (1.7) relating $\Sigma_{\infty}$ and $Q_{0}$. The next result
involves $\Sigma_{1}$ and $Q_{1}$. (We haven’t exp[ored to what extent the
boundary smoothness might be relaxed. We need to apply the Divergence
Theorem.)

###### Theorem 1

For (convex) domains $\Omega$ with Lipschitz boundary which is also piecewise
$C^{1}$, the following inequality is satisfied:

$\frac{A^{2}}{L}-\frac{(\Sigma_{\infty}-Q_{0})^{2}}{\Sigma_{1}}\leq Q_{1}.$
(2.1)

Recall that $\Sigma_{1}<0$, so both terms on the left-hand side are positive.

Proof. Considering ${\rm div}(u_{0}\nabla u_{\infty})$ the Divergence Theorem
gives

$Q_{0}=\int_{\Omega}\nabla u_{0}\,\cdot\,\nabla_{\infty}.$

Considering ${\rm div}(u_{\infty}\nabla u_{0})$ the Divergence Theorem gives

$\int_{\partial\Omega}u_{\infty}\frac{\partial u_{0}}{\partial
n}=-\Sigma_{\infty}+\int_{\Omega}\nabla u_{0}\,\cdot\,\nabla u_{\infty}.$

These combine to give

$\Sigma_{\infty}-Q_{0}=\int_{\partial\Omega}u_{\infty}\left(-\frac{\partial
u_{0}}{\partial n}\right).$

We now introduce an arbitrary constant $c$ and subtract $c\,A$ from both
sides, giving

$(\Sigma_{\infty}-Q_{0})-cA=\int_{\partial\Omega}(u_{\infty}-c)\,\left(-\frac{\partial
u_{0}}{\partial n}\right).$ (2.2)

We now use the Cauchy-Schwarz inequality on the right-hand side to give

$\left((\Sigma_{\infty}-Q_{0})-cA\right)^{2}\leq
Q_{1}\,\int_{\partial\Omega}(u_{\infty}-c)^{2},$

where we have used equation (3.1) for one of the integrals on the right-hand
side. Now, on using that the integral of $u_{\infty}$ around the boundary is
zero, we have

$\int_{\partial\Omega}(u_{\infty}-c)^{2}=-\Sigma_{1}+c^{2}L.$

Thus, for all real $c$

$\frac{\left((\Sigma_{\infty}-Q_{0})-cA\right)^{2}}{-\Sigma_{1}+c^{2}L}\leq
Q_{1}$ (2.3)

The function of $c$ on the left is clearly nonnegative and bounded, tending to
$A^{2}/L$ as c tends to both plus and minus infinity. It has two critical
points: the one making the function 0 is clearly the minimum. The other is at
$c=c_{*}$ where

$c_{*}=\frac{A\Sigma_{1}}{L(\Sigma_{\infty}-Q_{0})}.$

Subsitituting $c_{*}$ for $c$ in inequality (2.3) yields the result of the
theorem. $\square$

##  3 Tangential polygons

###  3.1 Geometric results

A tangential polygon, also known as a circumscribed polygon, is a convex
polygon that contains an inscribed circle (also called an incircle), a circle
that is tangent to each of the polygon’s sides. For any (convex) tangential
polygon, the area $A$, perimeter $L$ and inradius $\rho$ are related by

$A=\frac{1}{2}\rho L.$

We always choose the origin of our coordinate system to be at the incentre.
There is some literature on tangential polygons, e.g. [2, 74]. Any triangle is
a tangential polygon. Quadrilaterals which are tangential include kites and
hence rhombi.

Concerning $B$, defined in [69] and here at equation (1.3), another simple
identity for (convex) tangential polygons is

$B=\frac{L}{\rho},$

(and $B\geq{2\pi}$ with equality only for the disk, and for any triangle,
$B\geq{6\sqrt{3}}$): see [1].

The quantities $\Sigma_{\infty}$ and $\Sigma_{1}$ can be expressed in terms of
boundary moments $i_{2k}$ – moments about the incentre – defined by

$i_{2k}=\int_{\partial\Omega}(x^{2}+y^{2})^{k}.$

We remark that the Cauchy-Schwarz inequality for the integrals implies that
$i_{4}\geq{i_{2}^{2}/L}$.

For methods to calculate $i_{2k}$ in terms of the tangential polygon’s
inradius and angles, see [47].

###  3.2 The new inequality for tangential polygons

Since on the boundary $\partial\Omega$ of a tangential polygon we have
$x\cdot{n}=\rho$, equation (1.2) yields, for tangential polygons,

$Q_{0}=\frac{\rho}{4}\int_{\partial\Omega}\left(\frac{\partial u_{0}}{\partial
n}\right)^{2}.$ (3.1)

This enables us to rewrite the preceding theorem as follows.

###### Theorem 2

For any tangential polygon $Q_{0}$ satisfies the inequality, quadratic in
$Q_{0}$,

$\frac{A^{2}}{L}-\frac{(\Sigma_{\infty}-Q_{0})^{2}}{\Sigma_{1}}\leq\frac{4}{\rho}Q_{0}.$
(3.2)

In our application we treat inequality (3.2) as a quadratic inequality in
$Q_{0}$ and it is satisfied if

$Q_{0-}\leq Q_{0}\leq Q_{0+},$

where, with

$\delta=-\frac{\Sigma_{1}}{L}\left(2AL^{2}\Sigma_{\infty}-L^{3}\Sigma_{1}-A^{4}\right),$
$Q_{0\pm}=\frac{1}{A}\left(-L\Sigma_{1}+A\Sigma_{\infty}\pm\sqrt{\delta}\right).$

###  3.3 $\Sigma_{\infty}$ and $\Sigma_{1}$

For a tangential polygon

$u_{\infty}=c_{0}-\frac{1}{4}r^{2}\qquad{\rm where\ }r^{2}=x^{2}+y^{2},\ {\rm
and\ \ }c_{0}=\frac{i_{2}}{4L}.$ (3.3)

$c_{0}$ is such that the boundary integral is zero. This was noted at
equations (2.14), (2.15) of [44]. We find

$\displaystyle\Sigma_{\infty}$ $\displaystyle=$ $\displaystyle\frac{1}{16}\rho
i_{2}=\frac{Ai_{2}}{8L},$ (3.4) $\displaystyle\Sigma_{1}$ $\displaystyle=$
$\displaystyle-
c_{0}^{2}\,L+\frac{1}{2}c_{0}i_{2}-\frac{1}{16}i_{4}=\frac{1}{16}\left(\frac{i_{2}^{2}}{L}-i_{4}\right).$
(3.5)

We can now readily rewrite our inequality (3.2) in terms of the geometric
quantities $i_{2}$ and $i_{4}$. Using the expressions (3.4,3.5) in terms of
$i_{2}$, $i_{4}$ we find that the inequality of (3.2) is satisfied iff the
following inequality, quadratic in $Q_{0}$, is satisfied

$f(Q_{0}):=32AQ_{0}^{2}-\frac{4Q_{0}}{\rho}\left(2Ai_{4}-\rho
i_{2}^{2}+Ai_{2}\rho^{2}\right)+A\rho\,(Ai_{4}-\frac{3}{8}i_{2}^{2}\rho)\leq{0}\
.$ (3.6)

Inequalities on the domain functionals $A$, $\rho$, $i_{2}$ and $i_{4}$ can be
used to establish that both roots of $f(Q_{0})=0$ are positive. Denote the
smaller root by $Q_{0-}$ and the larger by $Q_{0+}$. The inequalities
$Q_{0-}\leq Q_{0}\leq Q_{0+}$ improve, for tangential polygons, some well
known inequalities such as the Polya-Szego inequality (1.4),

$Q_{0}\geq\frac{A^{2}}{4\ B_{\Omega}}(={\rm{\ for\ tangential\ polygons\
}}\frac{A^{3}}{2L^{2}}=\frac{\rho^{2}A}{8}).$

We now comment on the quadratic $f$. Consider first a disk radius 1

$A_{\odot}=\pi,\ \rho_{\odot}=1,\ i_{2,\odot}=i_{4,\odot}=2\pi,$ $None$

This agrees with that $Q_{0,\odot}=\pi/8$. Next consider an equilateral
triangle,

$A_{\Delta}=\sqrt{3},\ \rho_{\Delta}=\frac{1}{\sqrt{3}},\ i_{2,\Delta}=4,\
i_{4,\Delta}=\frac{16}{5},$ $None$

which is consistent with $Q_{0,\Delta}=\sqrt{3}/20$.

Consider next general tangential polygons. Define $Q_{B}=\rho^{2}A/8$, and
recall the Polya-Szego inequality $Q_{0}\geq Q_{B}$. We have

$f(Q_{B})=\frac{\rho^{2}A}{2}\left(\rho A-\frac{1}{2}\rho i_{2}\right)^{2}.$

Thus the inequality $Q_{0}\geq Q_{0-}$ improves on $Q_{0}\geq Q_{B}$. Consider
next the upper bound $Q_{0}\leq{I_{O}/4}$ where $I_{O}$ is the polar moment of
inertia about the incentre $O$, which can also be written
$Q_{0}\leq\Sigma_{\infty}$. Then $I_{O}=4\Sigma_{\infty}=\rho i_{2}/4$ and

$f(\frac{1}{16}\rho i_{2})=-\frac{1}{2}\left(\frac{i_{2}}{2}-\rho
A\right)\,\left(2Ai_{4}-\rho i_{2}^{2}\right).$

Since both terms in parentheses are positive, one concludes that $\rho
i_{2}/16\leq Q_{0+}$ so the bound $Q_{0}\leq Q_{0+}$ is weaker than the
earlier bound. Summarizing, we have

$Q_{B}=\frac{1}{8}\rho^{2}A\leq Q_{0-}\leq Q_{0}\leq\frac{1}{16}\rho i_{2}\leq
Q_{0+}.$

Using the calculations of $\Sigma_{\infty}$ and $\Sigma_{1}$ for isosceles
triangles with area $\sqrt{3}$ given in Part II we show in Figure 1 how the
new inequality compares with earlier inequalities. As another check we note
that perhaps the most studied isosceles triangle other than the equilateral is
the right isosceles triangle. Let $\alpha$ be the apex angle of the isosceles
triangle and $\sigma=\tan(\alpha/4)$ so $0<\sigma<1$. For this
$\sigma=\sqrt{2}-1\approx{0.414214}$ and for area $\sqrt{3}$ its torsional
rigidity is approximately 0.07827 (see [69]) which is, as it must be, larger
than $Q_{0-}$ which, at this $\sigma$, is 0.076511. However, at just 2.5%
difference, it is too close to the curve to plot usefully.

##  4 Conclusion, and open problems

There are many questions.

We do not know if inequality (2.1) is implied by some other known inequality
for torsional rigidity.

Inequality (3.2) becomes an equality for circular disks and equilateral
triangles. It may be that these are the only shapes which achieve this.

For isosceles triangles with a given area, as indicated in Figure 1, $Q_{0-}$
is maximized (and $Q_{0+}$ minimized) by the equilateral triangle. We believe
that this will be true for all triangles. What happens with quadrilaterals,
and more generally $n$-gons, is not known.

It would also be possible to check how well the inequality checks with
computed torsional rigidities for polygons, a few references being [32, 75,
79]. Perhaps checking for the regular polygons where the data is given in
Table 1 of Part II would be the easiest place to start.

To date the work has been on domains for which $u_{\infty}$ is a quadratic
polynomial. ([50] treats rectangles.) Outside this class of domains, there are
other domains for which all the functionals in inequality (2.1) can be found,
for example, the semi-circle has elementary function solutions for both
$u_{0}$ and $u_{\infty}$ and the various functionals can be found (using Maple
to sum the series).

## Structure of the remainder of this document

The preceding part of this document, called Part I from now on, has been
accepted, subject to revision, for publication in IMA Journal of Applied
Mathematics.
One referee was very positive and concluded:
“This paper too is very well written, clear, and easy to follow. I strongly
recommend publication in IMA Journal of Applied Mathematics”
The other referee wrote of the paper:
“the work was not put into context and background/other related studies were
not discussed …. I think it’d be good to discuss previous work on this topic,
so that it is clear what the new contribution of this work is. The derived
inequalities will be interesting once a discussion is given (e.g of other
inequalities which have appeared in the literature, why are these important
etc.). ”
The same referee also suggested:
“computing torsional rigidities for specific cases and comparing to other
studies”.
This supplement is intended to address this, while leaving the journal paper
short and focussed on the new research.
Numerics for the torsional rigidities of regular polygons and of isosceles
triangles (in the papers submitted to IMA) are repeated here in Parts I and
IIa. Additional numerics for rhombi are given near the end of Part III.

  * •

The earlier part of this document, Part I, is a preprint form of the IMA
paper. Some items addressing a referee’s concerns with the original form or
the paper are in Part Ib.

  * •

Part II concerns geometric matters relating to tangential polygons.
Part IIa is adapted from material used in a different context in the paper
[47]
Part IIb contains geometric items not in the IMA paper(s).
The first topics are related to $n$-gons including‘isoperimetric
inequalities’, when, with the number of vertices $n$ fixed, with some fixed
parameter (e.g. area) regular polygons optimize over some other parameter
(e.g. minimize the perimeter).
A second topic is inequalities, sometimes not involving $n$ or at least
allowing $n$ to range over positive integers, ‘circumgons’ and
‘circum-$n$-gons’, i.e. shapes in which part of the boundary is the disk with
radius the inradius. One of these is the ‘single-cap’, the convex hull of the
disk and a single point outside it: a circum-1-gon. Another is the ‘symmetric
double-cap’: a circum-2-gon. We will see these in connection with Blaschke-
Santalo diagrams.

  * •

In Part III we return to considerations of torsional rigidity. The emphasis is
on bounds for convex domains, in particular convex polygons especially
tangential polygons. Some sections are devoted to triangles, especially
isosceles, and tangential quadrilaterals, especially rhombi.

  * •

The remaining parts are only slightly connected to the torsion problem.
Part IV concerns replacing the Dirichlet b.c. with a Robin b.c..
Part V notes some other pde problems where tangential polygons are mentioned.

The treatment is very uneven. I have not checked the more advanced recent pde
papers. Some of the geometry papers cited in Part II are very elementary. The
suggestion (by Buttazzo) that I look at Blaschke-Santalo diagrams has led to
items at present poorly integrated with the study of my bound $Q_{0-}$ (with
just Part III §27 indicating one direction). I hope a later version of this
supplement will correct some of these defects.

Part Ib

## Numerics for the isosceles triangle

Numerics for the isosceles triangle were described earlier, but here is some
amplification.

Figure 1: For an isosceles triangle with area$\sqrt{3}$. $\sigma$ is tan of
1/4 of the apex angle. Blue is $Q_{B}$, red is the new lower bound $Q_{0-}$,
black is $Q_{\Delta}$, orange is $\rho i_{2}/16$, green is $Q_{0+}$.

Another lower bound on $Q_{0}$, as in [82], is that, amongst triangles with a
given inradius, the equilateral triangle has the smallest $Q_{0}$. Thus

$Q_{0}\geq Q_{\rm Sol}=\frac{9\sqrt{3}}{20}\rho^{4}.$ (4.1)

For isosceles triangles this lower bound improves on our $Q_{0-}$ when the
apex angle is slightly less than $\pi/3$. See Figure 2.

Figure 2: A plot, against $\sigma$, of the difference $Q_{0-}-Q_{\rm Sol}$,
the latter term defined in equation(4.1).

## Speculation on equality in inequality (3.2).

Inequality (3.2) becomes an equality for circular disks and equilateral
triangles. I don’t have a proof, but it might be that one only gets equality
for these shapes.

There is just one inequality used in the proof of the theorems. It is a
Cauchy-Schwarz Inequality and it will be an equality iff

$u_{\infty}-c_{*}={\rm const}\,\frac{\partial u_{0}}{\partial n},$

all the way around the boundary. The lhs is a quadratic function of $(x,y)$.
Consider next a genuine polygon, a tangential $n$-gon. Without loss of
generality consider a side parallel to the $x$-axis as the interval
$-1\leq{x}\leq{1}$, and $y=-h$, with the $n$-gon being in the half-plane
$\\{y>-h\\}$. The only possibility is

$\frac{\partial u_{0}}{\partial y}=-\frac{1}{4}(1-x^{2}),\qquad{\rm on\
}y=-h,\ \ -1<x<1,$

as the gradient of $u_{0}$ is zero at the corners. This has implications for
the values of $c_{0}$ in $u_{\infty}$ and $c_{*}$. Further information might
be obtained by considering sides adjoining the $y=-h$ side, with angles
$\alpha_{-}$ at $x=-1$ and $\alpha_{+}$ at $x=1$. One might get other
restrictions on the $c_{0}$ and $c_{*}$ values, and on the $\alpha\pm$.

Even if the disk and equilateral triangle are the only bounded domains for
which we get equality, it might be difficult to show this. If one considers an
infinite wedge as ‘tangential polygon’ this might be a counterexample.
Consider a wedge, apex at the origin and symmetric about the $x$-axis,
$\theta=0$. Suppose that the wedge has angle $\alpha$. Then

$u_{0,{\rm
wedge}}=-\frac{r^{2}}{4}\left(1-\frac{\cos(2\theta)}{\cos(\alpha)}\right)+{\rm
const}r^{\pi/\alpha}\cos(\pi\frac{\theta}{\alpha}),$

solves the torsion equation, has zero Dirichlet boundary data. Here

$\frac{\partial u_{0}}{\partial n}=\frac{1}{r}\frac{\partial
u_{0}}{\partial\theta}.$

It might be that we can set the const to 0 to see a counter example. (The
solution in a sector is given in [43].)

Part IIa: $\Sigma_{\infty}$ and $\Sigma_{1}$ for tangential polygons

## Abstract for Part IIa

Items useful in calculation for tangential polygons in general, and for
particular cases, are here extracted from [47].

##  5 Outline of Part IIa

In this part the functionals $\Sigma_{\infty}$ and $\Sigma_{1}$ are calculated
for various tangential polygons. The starting point is equation (2.14), (2.15)
from the 1993 paper [44] – repeated at appropriate times in this document –
which gives $u_{\infty}=c_{0}-\frac{1}{4}(x^{2}+y^{2})$ for tangential
polygons, with $c_{0}$ such that the boundary integral of $u_{\infty}$ is 0.

  * •

In §6 we find that starting from $u_{\infty}$ the functionals
$\Sigma_{\infty}$ and $\Sigma_{1}$ can be expressed in terms of various
moments of inertia.

  * •

In §7 we treat the circular disk and the equilateral triangle.

  * •

In §8 we study regular $n$-gons.

  * •

Any triangle is a tangential polygon.
In §9 we find the functionals for any isosceles triangle.

  * •

In §10. we note papers relevant to work on tangential quadrilaterals.

We will use ‘tangential polygon’ as defined before.

  * •

Genuine $n$-sided polygons for which every side is tangent to the incircle
(and hence for which the intersection of the boundary with the incircle is $n$
points, the points of tangency) will be called tangential $n$-gons.

  * •

When tangential polygon is the convex hull of $n$ points outside the incircle
and the incircle we will call it a circum-$n$-gon. (This is a slightly
different terminology than in [2].

Tangential $n$-gons are particular cases of circum-$n$-gons. The union over
all $n$ from 0 to $\infty$ of circum-$n$-gons gives all tangential polygons
(the $n=0$ case being considered as the disk). A circum-1-gon is also called a
1-cap, a circum-2-gon is also called a 2-cap: we will see these in Part IIb.

We use established terminology with a circum-4-gon being called a tangential
quadrilateral, a circum-6-gon a tangential hexagon, etc.

##  6 Tangential polygons

###  6.1 Geometric results

Here we continue from the basic geometric definitions and results given in
Part IIa §5.

There are various well-known or elementary observations:

  * •

Of the polygons with fixed perimeter and angles, the tangential polygon has
the greatest area.

  * •

Given two (convex) tangential polygons with the same incircle their
intersection is also a (convex) tangential polygon with the same incircle.

Papers concerning tangential polygons include [2, 74] (and many more
particularly concerning tangential $n$-gons for $3\leq{n}\leq{6}$ will be
given at appropriate places later in this document). There is a literature on
(convex) tangential polygons, one fact being that, considering the polygons as
linkages touching the incircle, if the number of sides is odd the polygon is
rigid but not if the number of sides is even. Entertaining as such facts are,
they do not appear to be relevant to our pde exercise.

We will need boundary moments $i_{2k}$ – moments about the incentre – defined
by

$i_{2k}=\int_{\partial\Omega}(x^{2}+y^{2})^{k},$

and the polar area moment about the incentre

$I_{2k}=\int_{\Omega}(x^{2}+y^{2})^{2}.$

(Caution. There are many results concerning moments about the centroid. For
example, as in [69], the moments about the centroid are minimized over
$n$-gons with a given fixed area by the regular $n$-gon. We haven’t checked in
general if it is the case that the moments about the incentre are minimized
over $n$-gons by the regular $n$-gon, but when $n=3$ and one minimizes over
isosceles triangles, the equilateral triangle is the minimizer.)

We remark that the Cauchy-Schwarz inequality for the integrals implies that
$i_{4}\geq{i_{2}^{2}/L}$ (and $I_{4}\geq{I_{2}^{2}/A}$).

####  6.1.1 $I_{2k}=\rho\,i_{2k}/(2k+2)$.

A result which we first noticed in two special cases is the following.

Result. For any tangential polygon $A=\rho\,L/2$, $I_{2}=\rho\,i_{2}/4$ and,
more generally, $I_{2k}=\rho\,i_{2k}/(2k+2)$.

Proof. The Divergence Theorem can be used to give a boundary integral which
equals $I_{2}$. Consider polar coordinates, radial coordinate $r$, and the
Laplacian of a function of $r$:

$\Delta u=\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial u}{\partial
r}.$

Also, for use in the following, for any tangential polygon $x.n=\rho$, here
$x$ being the vector to a point on $\partial\Omega$. To avoid using $x$ as a
vector, equally $re_{r}.n=\rho$, with $e_{r}$ the unit vector in the
$r$-direction.

With $u=r^{2}$ so $\Delta r^{2}=4$ the Divergence Theorem gives

$2\rho L=\int_{\partial\Omega}2re_{r}.n=\int_{\Omega}4=4A.$

Similarly, with $u=r^{4}$ so $\Delta r^{4}=16r^{2}$ the Divergence Theorem
gives

$4\rho i_{2}=\int_{\partial\Omega}(\nabla
r^{4}).n=\int_{\Omega}16\,r^{2}=16I_{2}.$

Alternative Proof. Start with a tangential polygon $T(1)$ origin at the
incentre, with radius 1. Define $T(\rho)=\rho\,T(1)$ with second area moment
$I_{2}(\rho)$ and second boundary moment $i_{2}(\rho)$. Along with $T(\rho)$
consider a similar scaled polygon $T(\rho+\Delta\rho)$ with $\Delta\rho$
small. Then

$I_{2}(\rho+\Delta\rho)-I_{2}(\rho)\sim\Delta\rho\,i_{2}(\rho)+O((\Delta\rho)^{2}).$

But, dimensionally, $i_{2}(\rho)=\rho^{3}\,i_{2}(1)$ and
$I_{2}(\rho)=\rho^{3}\,I_{2}(1)$. Taking limits in the displayed equation
gives the central part of

$4\rho^{3}I_{2}(1)=\frac{dI_{2}(\rho)}{d\rho}=i_{2}(\rho)=\rho^{3}i_{2}(1).$

This gives $i_{2}(1)=4I_{2}(1)$ and hence, more generally,
$I_{2}(\rho)=\rho\,i_{2}(\rho)/4$ as asserted.

The same methods give $I_{2k}=\rho\,i_{2k}/(2k+2)$, and the case $k=0$ is
$A=\rho\,L/2$.

####  6.1.2 Methods to calculate $i_{2k}$.

Denote with a bar just contributions from a vertical side, at $x=\rho$,
extending from $y=-\eta_{-}$ to $y=\eta_{+}$. Then

${\bar{i}}_{2k}=\int_{-\eta_{-}}^{\eta_{+}}(\rho^{2}+y^{2})^{k}\,dy.$

Hence

$\displaystyle{\bar{i}}_{0}$ $\displaystyle=$
$\displaystyle\eta_{+}+\eta_{-},$ $\displaystyle{\bar{i}}_{2}$
$\displaystyle=$
$\displaystyle\rho^{2}{\bar{i}}_{0}+\frac{1}{3}(\eta_{+}^{3}+\eta_{-}^{3}),$
$\displaystyle{\bar{i}}_{4}$ $\displaystyle=$
$\displaystyle-\rho^{4}{\bar{i}}_{0}+2\,\rho^{2}{\bar{i}}_{2}+\frac{1}{5}(\eta_{+}^{5}+\eta_{-}^{5}).$

Assuming the tangential polygon has $n$ sides, this gives

$\displaystyle{i}_{0}=L$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}(\eta_{k+}+\eta_{k-}),$ (6.1)
$\displaystyle{i}_{2}$ $\displaystyle=$
$\displaystyle\rho^{2}{i}_{0}+\frac{1}{3}\sum_{k=1}^{n}(\eta_{k+}^{3}+\eta_{k-}^{3}),$
(6.2) $\displaystyle{i}_{4}$ $\displaystyle=$
$\displaystyle-\rho^{4}{i}_{0}+2\,\rho^{2}{i}_{2}+\frac{1}{5}\sum_{k=1}^{n}(\eta_{k+}^{5}+\eta_{k-}^{5}).$
(6.3)

There is the geometrically evident fact that $\eta_{k+}=\eta_{k+1,-}$ as one
traverses from one of the polygon’s sides to the next.

The easiest case to consider is the regular $n$-gon, side
$s_{n}=2\rho\tau_{n}$ where $\tau_{n}=\tan(\pi/n)$. Then

$\displaystyle{i}_{0}=L$ $\displaystyle=$ $\displaystyle
n\,s_{n}=2n\rho\tau_{n},$ (6.4) $\displaystyle{i}_{2}$ $\displaystyle=$
$\displaystyle\rho^{2}{i}_{0}+\frac{n}{12}s_{n}^{3}=n\,s_{n}\left(\rho^{2}+\frac{1}{12}s_{n}^{2}\right)=\frac{2}{3}n\rho^{3}\tau_{n}(3+\tau_{n}^{2}),$
(6.5) $\displaystyle{i}_{4}$ $\displaystyle=$
$\displaystyle-\rho^{4}{i}_{0}+2\,\rho^{2}{i}_{2}+\frac{n}{80}s_{n}^{5}=n\,s_{n}\left(\rho^{4}+\frac{\rho^{2}s_{n}^{2}}{6}+\frac{s_{n}^{4}}{80}\right),$
(6.6) $\displaystyle=$
$\displaystyle\frac{2}{15}n\rho^{5}\tau_{n}\left(15+10\,\tau_{n}^{2}+3\,\tau_{n}^{4}\right).$
(6.7)

These had been calculated independently from first principles, as reported in
§8.1, before our observations concerning general tangential polygons.

We now consider how $\eta_{+}$ and $\eta_{-}$ might be found in terms of
vertex angles of the tangential polygon. Recall that in any tangential polygon
the angle bisector at any vertex passes through the incentre. Suppose side $k$
lies between vertices $k$ and $k+1$. Let the angle at vertex $k$ be
$\alpha_{k}$. (If there are $n$ sides the sum over all the $\alpha_{k}$ is
$(n-2)\pi$.) Then, with

$T_{k}=\frac{1}{\tan\frac{\alpha_{k}}{2}}=\tan(\frac{\pi-\alpha_{k}}{2}),$
(6.8) $\eta_{k-}={\rho}T_{k},\qquad\eta_{k+}=\rho T_{k+1}.$

Summing over $k$

$A=\rho^{2}{\sum\atop{k}}T_{k},\qquad L=2\rho{\sum\atop{k}}T_{k}.$ (6.9)

(This checks with our previous $A=\rho L/2$.) Another well-known identity is

$\frac{L^{2}}{4A}={\sum\atop{k}}T_{k}.$

Similarly

$\displaystyle{i}_{2}$ $\displaystyle=$
$\displaystyle\rho^{2}{i}_{0}+\frac{2\rho^{3}}{3}\sum_{k=1}^{n}T_{k}^{3}=2\rho^{3}\,\sum_{k=1}^{n}\left(T_{k}+\frac{T_{k}^{3}}{3}\right),$
(6.10) $\displaystyle{i}_{4}$ $\displaystyle=$
$\displaystyle-\rho^{4}{i}_{0}+2\,\rho^{2}{i}_{2}+\frac{2\rho^{5}}{5}\sum_{k=1}^{n}T_{k}^{5}$
(6.11) $\displaystyle=$ $\displaystyle
2\rho^{5}\,\sum_{k=1}^{n}\left(T_{k}+\frac{2T_{k}^{3}}{3}+\frac{T_{k}^{5}}{5}\right).$

###  6.2 $\Sigma_{\infty}$ and $\Sigma_{1}$

For a tangential polygon

$u_{\infty}=c_{0}-\frac{1}{4}r^{2}\qquad{\rm where\ }r^{2}=x^{2}+y^{2},$

and $c_{0}$ is such that the boundary integral is zero:

$c_{0}=\frac{i_{2}}{4L}.$ (6.12)

where $i_{2k}$ are the boundary moments defined previously. This was noted at
equations (2.14), (2.15) of [44]. Hence

$\displaystyle\Sigma_{\infty}$ $\displaystyle=$
$\displaystyle\frac{A\,i_{2}}{4L}-\frac{1}{4}I_{2}=\frac{1}{16}\rho
i_{2}=\frac{Ai_{2}}{8L},$ (6.13) $\displaystyle\Sigma_{1}$ $\displaystyle=$
$\displaystyle-
c_{0}^{2}\,L+\frac{1}{2}c_{0}i_{2}-\frac{1}{16}i_{4}=\frac{1}{16}\left(\frac{i_{2}^{2}}{L}-i_{4}\right),$
(6.14)

and the notation, as before, has $I$ for area moments, and $i$ for boundary
moments.

##  7 Equilateral triangle and disk

The results for these domains are well known: see e.g. [61]. For the unit disk

$A=\pi,\ L_{\odot}=2\pi,\ Q_{\odot,0}=\Sigma_{\odot,\infty}=\frac{\pi}{8},\
\Sigma_{\odot,1}=0,\ i_{\odot,2}=i_{\odot,4}=2\pi,\
I_{\odot,2}=\frac{\pi}{2}.$

For an equilateral triangle with side $2a=s_{3}$,

$A_{\Delta}=a^{2}\,\sqrt{3},\ L_{\Delta}=6a,\
Q_{\Delta,0}=\frac{\sqrt{3}a^{4}}{20}=\frac{A_{\Delta}^{2}}{20\sqrt{3}},\ $
$\Sigma_{\Delta,\infty}=\frac{a^{4}}{4\sqrt{3}}=\frac{A_{\Delta}^{2}}{12\sqrt{3}},\
\Sigma_{\Delta,1}=\frac{A_{\Delta}^{5/2}}{90\,\ 3^{1/4}}.$
$i_{\Delta,2}=\frac{4}{3}\,3^{1/4}\,A^{3/2},\
i_{\Delta,4}=\frac{16}{45}\,3^{3/4}\,A^{5/2},\
I_{\Delta,2}=\frac{\sqrt{3}}{9}\,A^{2}.$

For triangles and disks, both with area $\pi$, the St Venant isoperimetric
inequality is consistent with

$0.3927=Q_{\odot,0}=\frac{\pi}{8}>Q_{\Delta,0}=\frac{\pi^{2}}{20\sqrt{3}}=0.28491\
.$

The inequality for the $\Sigma_{\infty}$ is

$0.3927=\Sigma_{\odot,\infty}=\frac{\pi}{8}<\Sigma_{\Delta,\infty}=\frac{\pi^{2}}{12\sqrt{3}}=0.47485\
.$

##  8 Regular $n$-gons

###  8.1 General $n$

We denote the inradius by $\rho$, the side by $s$, area by $A$, perimeter by
$L$ and the angle at the centre subtended by a single side by $\gamma$. For
the (regular) $n$-gon, simple geometry gives $\gamma_{n}=2\pi/n$ so

$\frac{s_{n}}{2\rho_{n}}=\tau_{n}\qquad{\rm where\
}\tau_{n}=\tan(\frac{\gamma_{n}}{2})=\tan(\frac{\pi}{n}),$

and

$A_{n}=\frac{ns_{n}\rho_{n}}{2}=\frac{ns_{n}^{2}}{4\tau_{n}}=n\rho_{n}^{2}\tau_{n}.$

We will wish to specify the area (to be $\pi$), so we note

$s_{n}^{2}=\frac{4A_{n}\tau_{n}}{n}\qquad{\rm and\
}L_{n}=ns_{n}=2\sqrt{nA_{n}\tau_{n}}.$

The inradius $\rho_{n}$ occurs in some formulae, so we note

$\rho_{n}^{2}=\frac{A_{n}}{n\tau_{n}}.$

We are not aware of any simple formula for the torsional rigidity $Q_{0}(n)$
for a regular $n$-gon, but there have been many numerical studies (and
theoretical studies starting from Schwarz-Christoffel conformal mapping). Some
numerical results will be given for particular instances later.

Formulae for $i_{0}=L$, $i_{2}$ and $i_{4}$ have been presented earlier at
equations (6.4), (6.5) and (6.7). In view of their central role and the
detailed algebraic manipulations in their derivation we record here an
independent derivation. Using polar coordinates, and considering the side with
$x=\rho_{n}=r\cos(\theta)$ for which the polar angle at the centre lies
between $\theta=-\pi/n$ and $\theta=\pi/n$,

$i_{2}(n)=n\int_{-\pi/n}^{\pi/n}\left(\frac{\rho_{n}}{\cos(\theta)}\right)^{2}\,\frac{dy}{d\theta}\,d\theta,$

in which $y=\rho_{n}\tan(\theta)$. Thus

$\displaystyle i_{2}(n)$ $\displaystyle=$ $\displaystyle
n\rho_{n}^{3}\int_{-\pi/n}^{\pi/n}\frac{1}{\cos(\theta)^{4}}\,d\theta,$
$\displaystyle=$
$\displaystyle\frac{2\,n\rho_{n}^{3}\,\left(1+2\cos(\frac{\pi}{n})^{2}\right)\,\sin(\frac{\pi}{n})}{3\,\cos(\frac{\pi}{n})^{3}},$
$\displaystyle=$
$\displaystyle\frac{2}{3}n\rho_{n}^{3}\tau_{n}(3+\tau_{n}^{2}),$
$\displaystyle=$
$\displaystyle\frac{2}{3}\sqrt{\frac{A^{3}}{n\tau_{n}}}(3+\tau_{n}^{2}).$

Similarly

$\displaystyle i_{4}(n)$ $\displaystyle=$ $\displaystyle
n\rho_{n}^{5}\int_{-\pi/n}^{\pi/n}\frac{1}{\cos(\theta)^{6}}\,d\theta,$
$\displaystyle=$
$\displaystyle\frac{2\,n\rho_{n}^{5}\,\left(3+4\cos(\frac{\pi}{n})^{2}+8\cos(\frac{\pi}{n})^{4}\right)\,\sin(\frac{\pi}{n})}{15\,\cos(\frac{\pi}{n})^{5}},$
$\displaystyle=$
$\displaystyle\frac{2}{15}n\rho_{n}^{5}\tau_{n}(15+10\tau_{n}^{2}+3\tau_{n}^{4}).$

The area moment of inertia $I_{2}(n)$ is similarly calculated from that of the
isosceles triangle with apex at the origin. Before noting the general relation
at equation (6.13) a short calculation – just for regular polygons –
established that, for a regular polygon,

$I_{2}(n)=\frac{\rho_{n}}{4}\,i_{2}(n).$ (8.1)

Using equations (6.4), (6.5) and (6.7), equation (6.13) becomes

$\Sigma_{\infty}(n)=i_{2}(n)\,\left(\frac{A}{4L_{n}}-\frac{\rho_{n}}{16}\right)=\frac{A^{2}\,(3+\tau_{n}^{2})}{24n\tau_{n}}.$
(8.2)

(From (8.2), $\Sigma_{\infty}>\Sigma_{\odot,\infty}=Q_{\odot,0}=\pi/8$ when
$A=\pi$.) Also

$\Sigma_{1}(n)=-\frac{1}{16L}\left(Li_{4}-i_{2}^{2}\right)=-\frac{1}{90}\sqrt{\frac{A^{5}\,\tau_{n}^{5}}{n^{3}}}.$
(8.3)

The original motivation for assembling this particular list of quantities was
that they were needed for a bound associated with slip flow down a pipe with
cross-section $\Omega$: see [47]. Numeric values for the torsional rigidity
are available in several references. The entries for $J/A^{2}$, where
$J=4Q_{0}$, in the following table are taken from [32]. See also [75] and, for
$n=3,\ 4$ and $6$, also [69]. In Table 1 we take our polygons to have area
$\pi$. In Table 2 we take our polygons to have circumradius 1.

$n$ | $4Q_{0}/A^{2}$ | $Q_{0}$ | $L$ | $\Sigma_{\infty}$ | $-\Sigma_{1}$
---|---|---|---|---|---
3 | 0.11547 | 0.28492 | 8.0806 | 0.47485 | 0.14769
4 | 0.14058 | 0.34687 | 7.0898 | 0.41123 | 0.024296
5 | 0.14943 | 0.36870 | 6.7565 | 0.39936 | 0.007822
6 | 0.15340 | 0.37850 | 6.5978 | 0.39571 | 0.003349
7 | 0.15546 | 0.38358 | 6.5086 | 0.39426 | 0.001689
8 | 0.15664 | 0.38649 | 6.4530 | 0.39359 | 0.0009485
9 | 0.15736 | 0.38827 | 6.4159 | 0.39325 | 0.0005754
10 | 0.15783 | 0.38943 | 6.3899 | 0.39306 | 0.0003699
11 | 0.15815 | 0.39022 | 6.3709 | 0.39294 | 0.0002489
12 | 0.15837 | 0.39076 | 6.3566 | 0.39287 | 0.0001738
$\infty$ | 0.15915 | 0.3927 | 6.2832 | 0.3927 | 0
Table 1: Regular polygons with area $\pi$

$n$ | $4Q_{0}/A^{2}$ | $A_{n}$ | $Q_{0}$ | $L_{n}$ | $\Sigma_{\infty}$ | $-\Sigma_{1}$ | $\rho_{n}$
---|---|---|---|---|---|---|---
3 | $\sqrt{3}/15$ | $3\sqrt{3}/4$ | $9\sqrt{3}/320$ | $3\sqrt{3}$ | $3\sqrt{3}/64$ | $3\sqrt{3}/320$ | $1/2$
| 0.11547 | 1.2990 | 0.0487 | 5.1962 | 0.0812 | 0.0162 | 0.5
4 | 0.14058 | $2$ | 0.14058 | $4\sqrt{2}$ | $1/6$ | $\sqrt{2}/180$ | $1/\sqrt{2}$
|  | 2 |  | 5.6569 | 0.1667 | 0.007857 | 0.707107
6 | 0.15340 | $3\sqrt{3}/2$ | 0.2589 | $6$ | $5\sqrt{3}/32$ | $1/480$ | $\sqrt{3}/2$
|  | 2.5981 |  | 6 | 0.270633 | 0.0020833 | 0.866025
$n$ |  | $\frac{n}{2}\sin(\frac{2\pi}{n})$ |  | $2n\sin(\pi/n)$ | (8.2) | (8.3) | $\cos(\pi/n)$
$\infty$ | $1/(2\pi)$ | $\pi$ | $\pi/8$ | $2\pi$ | $\pi/8$ | 0 | 1
| 0.180043 |  |  |  |  | 0 | 1

Table 2: Regular polygons with circumradius 1

###  8.2 The square, $n=4$

For a square with side $s_{4}$,

$A_{\square}=s_{4}^{2},\ L_{\square}=4s_{4},\ I_{\square,2}=\frac{A^{2}}{6},\
\rho_{4}=\frac{1}{2}s_{4},\ i_{\square,2}=\frac{4}{3}s_{4}^{3}.$

Hence, consistent with equation (6.13), we have

$\Sigma_{\square,\infty}=\frac{s_{4}^{4}}{24}=\frac{A_{\square}^{2}}{24}.$

For a square and a disk each of area $\pi$

$0.3927=\Sigma_{\odot,\infty}=\frac{\pi}{8}<\Sigma_{\square,\infty}=\frac{\pi^{2}}{24}=\
0.41123.$

Calculation for a rectangle reported in [50] gives (with $a=b$)

$\Sigma_{\square,1}=-\frac{s_{4}^{5}}{720}=-\frac{A_{\square}^{5/2}}{720}.$

##  9 Triangles, especially isosceles

###  9.1 Geometric preliminaries and checks

####  9.1.1 $A$, $L$, $\Sigma_{\infty}$ and $\Sigma_{1}$ for isosceles
triangles

Consider the isosceles triangle whose incentre is at the origin, whose base is
length $2a$ and whose apex angle is $\alpha$. Denote by $\rho$ the inradius of
the triangle. The triangle’s vertices are at $(a,-\rho)$, $(0,h-\rho)$ and
$(-a,-\rho)$. The case where $h=a\sqrt{3}$, $\rho=h/3$ corresponds to an
equilateral triangle.

The area of the triangle is denoted by $A$. The vertex angle is $\alpha$.
Change the notation, what was formerly denoted $T_{k}$, we will denote
$T_{A}$, $T_{B}$ and $T_{C}$. It is convenient to define

$\sigma=\tan(\frac{\alpha}{4}).$

We have

$T_{A}=\tan(\frac{\pi-\alpha}{2})=\frac{1-\sigma^{2}}{2\sigma},\
T_{B}=T_{C}=\tan(\frac{\pi+\alpha}{4})=\frac{1+\sigma}{1-\sigma}.$

We remark though that an alternative parameter, alternative to $\sigma$, is
$a$, $2a$ being the length of the base. Then

$\displaystyle\frac{h}{a}$ $\displaystyle=$
$\displaystyle\cot(\frac{\alpha}{2})=\frac{1-\sigma^{2}}{2\sigma},$
$\displaystyle\frac{\rho}{a}$ $\displaystyle=$
$\displaystyle\tan(\frac{\pi-\alpha}{4})=\frac{1-\sigma}{1+\sigma},\qquad\rho=\frac{2A}{L},$

and, using equation (6.9) or directly,

$\displaystyle A$ $\displaystyle=$ $\displaystyle
a\,h=a^{2}\,\frac{1-\sigma^{2}}{2\sigma},$ $\displaystyle L$ $\displaystyle=$
$\displaystyle
2\left(a+\sqrt{a^{2}+h^{2}}\right)=a\frac{(1+\sigma)^{2}}{\sigma},$
$\displaystyle B$ $\displaystyle=$
$\displaystyle\frac{L}{\rho}=\frac{L^{2}}{2A}=\frac{(1+\sigma)^{3}}{\sigma(1-\sigma)}.$

There are several identities relating these geometric quantities. A relation
between the area, perimeter, and base is

$4La^{3}-L^{2}a^{2}+4A^{2}=0.$ (9.1)

This is consistent with the geometrically obvious

$L=2\left(a+\sqrt{a^{2}+(\frac{A}{a})^{2}}\right),$

which can be regarded as an example of using $a$ as a parameter. ($L(a)$ is a
positive convex function with, when $A=\sqrt{3}$ a minimum of 6 when $a=1$.)

Less immediately useful in this paper is the circumradius $R_{V}$. The radius
of the circumcircle (circle through the 3 vertice) is:

$R_{V}={\frac{a\,(a^{2}+(\frac{A}{a})^{2})}{2A}}.$

The centre of the circle lies on the symmetry axis of the triangle, this
distance below the apex.

An isosceles triangle with given area $A$ is uniquely determined if $\sigma$
is given between 0 and 1 or if $a$ is given between 0 and infinity, in the
latter case with

$\sigma=-\frac{A}{a^{2}}+\sqrt{\frac{A^{2}}{a^{4}}+1}.$

There are various formulae for the inradius $\rho$ including

$\displaystyle\frac{\rho}{a}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{a^{2}+h^{2}}-a}{h},$ $\displaystyle=$
$\displaystyle\frac{L-4a}{2h}.$

Result. At fixed area, $B$ and $L$ are positive convex functions of $\sigma$
for $0<\sigma<1$, each having its minimum at $\sigma=2-\sqrt{3}$. Over
isosceles triangles with given area, the equilateral triangle minimizes each
of $B$ and $L$.
At fixed area, $\rho$ is a positive concave function of $\sigma$ for
$0<\sigma<1$, having its maximum at $\sigma=2-\sqrt{3}$. Over isosceles
triangles with given area, the equilateral triangle maximizes $\rho$.

Integration gives, for the area moment,

$I_{2}(A,\sigma)=\frac{A^{2}}{12}\,\frac{(1-\sigma)^{6}+12\sigma^{2}(1-\sigma)^{2}+16\sigma^{3}}{\sigma(1-\sigma)(1+\sigma)^{3}}.$

Result. $I_{2}(A,\sigma)$ is a convex function of $\sigma$ for $0<\sigma<1$
with its minimum at $\sigma=2-\sqrt{3}$. Over isosceles triangles with given
area, the equilateral triangle minimizes the area moment of inertia about the
incentre.

There are various identities for $I_{2}$, e.g.

$4I_{2}-\frac{1}{6}AL^{2}+\frac{8A^{3}}{La}-\frac{16A^{3}}{L^{2}}+2Aa^{2}=0.$
(9.2)

Eliminating $L$ gives

$36a^{4}AI_{2}^{2}-12a^{2}(12a^{8}+11a^{4}A^{2}+A^{4})I_{2}+A(24a^{12}+33A^{2}a^{8}+6A^{4}a^{4}+A^{6})=0.$

ToDo. The discriminant of the quadratic for $I_{2}$ is nice, and the 2
solutions for $I_{2}$ are reasonably simple (and possibly tidier than the
expressions in $\sigma$). However surely only one is relevant, which one?
Might $i_{2}=I_{2}/(4\rho)$ (with $\rho=a^{2}(L-4a)/(2A)$) be neater than
$I_{2}$. See if there is an equation like (9.1) involving just $i_{2}$, $A$
and $a$. Equation (9.1) involves $i_{0}$, $A$ and $a$. If there is, it might
be that $i_{2}$ considered as a function of $a$ might be tidier than it is as
a function of $\sigma$.

For the boundary moment we split the integral into the part over the base, and
over another of the sides:

$i_{2k}=i_{2k}(base)+2\,i_{2k}(side).$

Once again, as in equation (8.1), we find, at $k=1$,

$I_{2}=\frac{\rho}{4}\,i_{2}.$

Result. At fixed area, $i_{2}$ is a positive convex functions of $\sigma$ for
$0<\sigma<1$, with its minimum at $\sigma=2-\sqrt{3}$. Over isosceles
triangles with given area, the equilateral triangle minimizes $i_{2}$ the the
boundary moment about the incentre.

Equation (6.13) becomes, with $i_{2}$ calculated either directly or from
(6.10),

$\displaystyle\Sigma_{\infty}(A,\sigma)$ $\displaystyle=$ $\displaystyle
i_{2}\,\left(\frac{A}{4L}-\frac{\rho}{16}\right)=\frac{1}{16}\rho
i_{2}=\frac{1}{4}I_{2},$ (9.3) $\displaystyle=$
$\displaystyle\frac{A^{2}}{48}\,\frac{((1-\sigma)^{6}+12\sigma^{2}(1-\sigma)^{2}+16\sigma^{3})}{\sigma(1-\sigma)(1+\sigma)^{3}}.$

Restating the preceding Result for $I_{2}$:
Result. $\Sigma_{\infty}(A,\sigma)$ is a convex function of $\sigma$ for
$0<\sigma<1$ with its minimum at $\sigma=2-\sqrt{3}$. Over isosceles triangles
with given area, the equilateral triangle minimizes $\Sigma_{\infty}$.

The formulae for $i_{4}$ and $\Sigma_{1}$ are more elaborate. See (6.11).
Define

$p_{1}(\sigma)=(1-\sigma)^{12}+9\sigma(1-\sigma)^{10}-40\sigma^{3}(1-\sigma)^{6}+144\sigma^{5}(1-\sigma)^{2}+256\sigma^{6}.$

One can show that $p_{1}$ is positive on $0\leq\sigma\leq{1}$.
The negative quantity $\Sigma_{1}$ is found as in (6.14):

$\displaystyle\Sigma_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{16}\left(\frac{i_{2}^{2}}{L}-i_{4}\right),$ (9.4)
$\displaystyle=$
$\displaystyle-\frac{1}{360}\,\frac{A^{3}}{L\,(\sigma(1-\sigma)(1+\sigma))^{3}}\
p_{1}(\sigma).$

Our main test cases are the equilateral triangle which has $\sigma=2-\sqrt{3}$
and the right isosceles triangle which has $\sigma=\sqrt{2}-1$. (Numerical
values for the torsional rigidities of other isosceles triangles are
available, for example in [67].)

It would, of course, be possible to produce tables, as done in §8.1 in the
different context of regular polygons, for a range of vertex angles for the
isosceles triangles. The starting point for this would be existing results for
$Q_{0}$, combined with our formulae for $L$, $\Sigma_{\infty}$ (9.3) and
$\Sigma_{1}$ (9.4).

###  9.2 The right isosceles triangle

$\alpha$ | $4Q_{0}/A^{2}$ | $A_{n}$ | $Q_{0}$ | $L_{n}$ | $\Sigma_{\infty}$ | $-\Sigma_{1}$
---|---|---|---|---|---|---
$\pi/3$ | $\sqrt{3}/15$ | $3\sqrt{3}/4$ | $9\sqrt{3}/320$ | $3\sqrt{3}$ | $3\sqrt{3}/64$ | $3\sqrt{3}/320$
$\rho=1/2$ | 0.11547 | 1.2990 | 0.0487 | 5.1962 | 0.0812 | 0.0162
$\pi/2$ | 0.10436 | $1$ | 0.02609 | $2+2\sqrt{2}$ | $(3-2\sqrt{2})/3$ | $(131-91\sqrt{2})/90$
$\rho=\sqrt{2}-1$ |  | 1 |  | 4.8284 | 0.0572 | 0.0256285

Table 3: Isosceles triangles with circumradius 1

##  10 Tangential quadrilaterals, especially kites and rhombi

A quadrilateral is tangential if and only if the sums of lengths of each pair
of opposite sides are equal. Examples of tangential quadrilaterals are the
kites, which include the rhombi, which in turn include the squares. (A
bicentric kite is an orthogonal kite: a bicentric rhombus is a square.)
Torsional rigidities have been found numerically, for rhombi in [70, 79]. The
other quantities occuring in the lower bound $R$ are easily found.

For example, the relevant quantities for a rhombus are as follows. Consider a
rhombus with inradius $\rho$, area $A$. Let $\alpha$ be the angle at an acute
vertex. The points of tangency of the incircle with the sides of the rhombus
divide each side into a smaller part $\eta_{-}$ and a larger part $\eta_{+}$.
Denote $\tan(\alpha/2)$ by $\tau$. Then $\eta_{-}=\rho\tau$ and
$\eta_{+}=\rho/\tau$. From equation (6.1)

$L=i_{0}=4\rho\left(\tau+\frac{1}{\tau}\right),\quad A=\frac{1}{2}\rho
L=2\rho^{2}\left(\tau+\frac{1}{\tau}\right),\quad\rho=\sqrt{\frac{A}{2\left(\tau+\frac{1}{\tau}\right)}}.$

(At fixed $A$, $L$ is minimized for the square, $\tau=1$.) We have, from
(6.2,6.3),

$\displaystyle i_{2}$ $\displaystyle=$
$\displaystyle\rho^{3}\left(4\left(\tau+\frac{1}{\tau}\right)+\frac{4}{3}\left(\tau^{3}+\frac{1}{\tau^{3}}\right)\right),$
$\displaystyle i_{4}$ $\displaystyle=$
$\displaystyle\rho^{5}\left(4\left(\tau+\frac{1}{\tau}\right)+\frac{8}{3}\left(\tau^{3}+\frac{1}{\tau^{3}}\right)+\frac{4}{5}\left(\tau^{5}+\frac{1}{\tau^{5}}\right)\right).$

These check, in the case $\tau=1$ with the quantities given in §8.2. Equations
(6.13,6.14) give $\Sigma_{\infty}$ and $\Sigma_{1}$ in terms of $A$, $L$,
$i_{2}$ and $i_{4}$. The results for general rhombi are used in Part IIb
§19.5.2 and in Part III §26.

There are many geometric results concerning tangential quadrilaterals. A
tangential quadrilateral is bicentric if and only if its inradius (hence area)
is greater than that of any other tangential quadrilateral having the same
sequence of side lengths. There may be similar results for some other domain
functionals.

Amongst all quadrilaterals with a given area that which
minimizes perimeter $L$,
maximizes $Q_{0}$ (or similarly $\dot{r}$)
is square: see [69] p159. The proof in [69] involves symmetrisation, with
kites to rhombi to rectangles then kites, etc.. Perhaps because of curiosity
on how the successive symmetrisations performed there have been numerical
studies of the torsion problem for rectangles, kites and rhombi. For rhombi an
early example is [70].

Part IIb: More geometry for tangential polygons

## Abstract for Part IIb

Further items on tangential polygons, additional to those in Part IIa (which
are taken from [47]), are collected here. In particular some Blaschke-Santalo
diagrams for some geometric functionals are presented.

##  11 Outline of Part IIb

In as much as the main focus of these notes should be the lower bound $Q_{0-}$
of Part I, we remark that Parts IIa and IIb only provide results on the
geometric quantities entering the formula for $Q_{0-}$, namely $\rho$, $A$,
$L$, $i_{2}$, $i_{4}$. We defer further treatment of $\Sigma_{1}$ and $Q_{0-}$
to Parts III and IV.

There are two main, and different, sorts of geometries.

  * •

One, especially prominent in §19, especially §19.2 and §19.4, involves general
tangential polygons, convex circumgons. (When we count the extreme points
outside the incircle these are, if $n$ points, called circum-$n$-gons.)

  * •

The other concerns genuine $n$-gons. Their boundaries consist solely of
straight line segments. The hope is that one can show that regular $n$-gons
optimize appropriate domain functionals over subsets of, sometimes all,
tangential $n$-gons. The hope is sometimes realized, an easy example (in
§16.1) being as follows:
Result.Let $\Omega_{0}$ be a tangential $n$-gon with inradius $\rho$, and
vertex angles $\alpha_{k}$. Let $\Omega_{1}$ be a tangential $n$-gon with the
same inradius $\rho$ and the same vertex angles except that vertices
$\alpha_{i}$ and $\alpha_{j}$ are each replaced by their average
$(\alpha_{i}+\alpha_{j})/2$. Then each of $L$, $A$, $i_{2}$, $i_{4}$ and
$d_{O}$ are reduced in the change from $\Omega_{0}$ to $\Omega_{1}$, strictly
so if $\alpha_{i}\neq{\alpha_{j}}$.
The ‘easy’ above refers to the proof. It wasn’t immediately obvious to this
author before the proof. By way of contrast, the following seems almost self-
evident.
Corollary.Amongst all tangential $n$-gons with a fixed inradius, the regular
$n$-gon minimizes each of $L$, $A$, $i_{2}$, $i_{4}$ and $d_{O}$.
We have yet to check the possibility that $Q_{0-}$ behaves, in this respect,
the same as [82] Theorem 1 gives for $Q_{0}$. Discussion of this is defered to
Part III.

Here is an outline of this part.

  * •

In §12 we consider operations involving tangential polygons.

  * •

In §13 we indicate how we coded to construct tangential polygons in order to
later compute domain functionals for them.

  * •

In §14 considerations of ‘duality’ direct the study.

  * •

In §15 we collect a somewhat miscellaneous set of inequalities and geometric
facts.

  * •

In §16 we consider triangles: in §17 tangential quadrilaterals.

  * •

In §18 we note a few facts concerning bicentric polygons. All triangles are
bicentric. All regular polygons are bicentric.

  * •

In §19 we present some information about Blaschke-Santalo diagrams for some
geometric functionals.

Blaschke-Santalo results can sometimes lead in to proving isoperimetric
results. Here is the style of an example with $\cal F$ some domain functional.

  * •

Amongst triangles with fixed $\rho$ and $L$ that which $<$ optimizes $\cal F$
$>$ is $<$ squat $|$ tall $>$ isosceles.

  * •

Amongst …isosceles triangles at fixed $A=\rho L/2$ that which $<$ optimizes
$\cal F$ $>$ is equilateral.

As an example consider $d_{O}$ (defined and treated extensively in §19). At
fixed $\rho$ and $A=\rho L/2$ squat isosceles triangles minimize $d_{O}$. With
this preliminary, when considering triangles with given $A$, minimizing
$d_{O}$ we need only consider isosceles triangles with that area. (For
$Q_{0-}$ I have in Part I seen that, amongst isosceles triangles with a given
area that which maximizes $Q_{0}$ is equilateral.) See §16 for more formal
treatment, but for now the following very informal discussion might make the
second step plausible. Consider now squat isosceles triangles with given
incentre and its base vertices on the circle radius $d_{O}$. It is eminently
plausible that increasing the inradius of this family of triangles will
increase the area, suggesting that at fixed $d_{O}$ one gets maximum $A$.
Conversely one expects at fixed $A$ to get minimum $d_{O}$ at the equilateral
triangle.

##  12 Transformations involving tangential polygons

Changing scale by some factor $t$ changes a tangential polygon with inradius
$\rho$ to one with inradius $t\rho$. Mostly we fix the inradius, and always
have the origin of coordinates at the centre of the incircle. In this
situation, as we have already noted, in Part IIa §6.1 that given two (convex)
tangential polygons with the same incircle their intersection is also a
(convex) tangential polygon with the same incircle. (For tangential $n$-gons
the number of vertices could increase.)

###  12.1 Convex $n$-gon to tangential $n$-gon

Given a convex $n$-gon, and hence its sequence of angles, all less than $\pi$,
then one can define a tangential $n$-gon with the same sequence of angles and
the same area. This transformation reduces the perimeter. (See [2].)

###  12.2 Tangential $n$-gons to tangential $m$-gons, $m\geq{n}$

As a particular case of the intersection of tangential polygons being
tangential polygon we mention “corner cutting”. Given a tangential polygon
$\Omega$ and a half-plane $H$ containing the incircle of $\Omega$, then
$H\cap{\Omega}$ is a tangential polygon. Both $L$ and $A$ are decreased while
$\rho=2A/L$ stays constant. The number of vertices increases.

Let $\Omega$ be a tangential polygon. Let $\sigma(l,.)$ be reflection through
a line $l$ through the origin. Then $\sigma(l,\Omega)$ is a tangential polygon
and so is its intersection with $\Omega$.

In the case of tangential $n$-gons, the numbers of vertices changes. For
example, if $\Omega$ is an equilateral triangle and $l$ is parallel to a side,
the intersection is a hexagon.

###  12.3 Tangential $n$-gons to tangential $m$-gons, $m\leq{n}$

Begin with a tangential polygon with $n\geq{4}$ vertices. Moving a tangency
point to an adjacent tangency point will result in an edge disappearing. While
the inradius stays the same, the new tangential polygon might not be bounded
with an example of this having the starting point as a square.

###  12.4 Permutations of angles of tangential $n$-gons

If one permutes the entries of a sequence of angles or a tangential polygon,
keeps the inradius the same, one has another tangential polygon with the same
$\rho$, $A$, $L$, $i_{2}$, $i_{4}$, $d_{O}$. As an example consider reflection
about any ‘diagonal’: the incentre moves, but the reflected polygon is
tangential.

###  12.5 Tangential $2m$-gons to 2-special $2m$-gons

Another transformation which at least takes tangential quadrilaterals to
tangential quadrilaterals is $m$-descendant mapping: see [92]. This paper also
defines a $n$-gon with $n=2m$ even to be 2-special if the sum of lengths even-
indexed sides is equal to that of the odd-indexed sides. Also the 2-descendant
map of a $n$-gon with sides $s_{{\rm in},j}$ is the $n$-gon with sides

$s_{{\rm out},j}=\frac{1}{2}(s_{{\rm in},j}+s_{{\rm in},j+1}).$

The 2-descendant map of any 2-special $2m$-gon is 2-special. In particular the
2-descendant map of a tangential quadrilateral is a tangential quadrilateral.
Repeated application of 2-descendant maps to an initial tangential $2m$-gon
would take one ever closer to a regular $2m$-gon. We remark that the doubly
stochastic circulant matrix $\frac{1}{2}M(n)$, defined in §15.1 is here
applied to ${\mathbf{s}}_{\rm in}$. The effect of many successive operations
with the map leading to the regular $n$-gon corresponds to the fact that the
matrix powers tend to the $1/n$ times the matrix $E$ all of whose entries are
1:

$\left(\frac{1}{2}M(n)\right)^{k}\rightarrow\frac{1}{n}E(n)\qquad{\rm as\ \
}k\rightarrow\infty.$

ToDo. Check out guess that applying a 2-descendant map to a tangential hexagon
may not lead to tangential hexagon. (It is known that being 2-special is
necessary but not sufficient for a hexagon to be tangential.)

One can also consider $2m$-gons as linkages. Again a tangential $2m$-gon can
move as a linkage to another 2-special configuration. For tangential
quadrilaterals the linkage remains, when convex, a tangential quadrilateral
(but this will not be the case for 6-gons).

###  12.6 Tangential $n$-gons to Tangential $n$-gons, preserving $\rho$

Figure 3: Diagram for ‘tilting transformation’

A transformation, called a ‘tilting transformation’ is now defined. In this
context refer to Figure 3. A side is ‘tilted’, its tangent contact point
moved, so that it becomes parallel to the line joining the contact points of
adjacent sides. All tangent points except one remain fixed. Suppose the
initial configuration is as shown in the figure. The red and green tangent
lines remain fixed, but suppose the tangent point of the black side is
considered movable. The fixed tangent points either side of the movable one
are at

$\zeta_{-}=(-X,h),\qquad\zeta_{+}=(X,h).$

The coordinates of the movable tangent point might be taken as

$\zeta=\rho\left(\frac{1-t^{2}}{1+t^{2}},\frac{2t}{1+t^{2}}\right)\ \ {\rm\
and\ w.l.o.g.\ }\rho=1.$

The area $A(t)$ of the quadrilateral obtained from intersecting the tangential
$n$-gon with the half-space $\\{y>h\\}$ can be found, as can the length $L(t)$
of the three line segments in the upper part of its perimeter. Finding the
formulae for $A(t)$ and $L(t)$ is not too onerous, with one check being that
$2A(t)-L(t)$ is independent of $t$. Both $A(t)$ and $L(t)$ are minimized at
$t=1$ which has the topmost (black) tangent line parallel to $y=h$ and the two
consecutive angles of the tangential $n$-gon equal.

Repeated application of the tilting transformation, for varying sides, will,
in the limit, get one to a tangential $n$-gon with all angles equal. If a
tangential $n$-gon has all angles equal it is regular. Thus, amongst all
tangential $n$-gons with radius $\rho$, that which has the smallest area (and
perimeter $L=2A/\rho$) is the regular $n$-gon.

Similarly, amongst all tangential $n$-gons with radius $\rho$, that which has
the smallest $d_{O}$ is the regular $n$-gon.

The ‘tilting transformation’ is easy to visualize geometrically. Less obvious
is that at fixed $\rho$ one can reduce $A$, $L$, $i_{2}$, $i_{4}$, $d_{O}$ by
averaging any two angles (not necessarily adjacent angles as above). The
reduction of $L$ is a consequence of the convexity of the $\cot$ function on
$(0,\pi/2)$, the quantities $T_{k}$ defined in Part IIa equation 6.8) and the
representation of $L$ as the sum of the $T_{k}$. The other quantities $i_{2}$,
etc. are treated similarly. More details are given in §16.1.

###  12.7 Circum-$n$-gons to circum-$m$-gons, $m\leq{n}$

Let $\Omega$ be a tangential polygon with inradius $\rho_{0}$ and with the
maximum distance of the boundary to the origin $d_{O}$. Then, for any ball $B$
centred at the origin, the convex hull ${\rm conv}(B\cup{\Omega})$ is a
tangential polygon.
ToDo. Prove this and investigate the behaviour of $L/\rho$ as the radius of
$B$ increases from $\rho_{0}$ to $d_{O}$.

###  12.8 Minkowski sums

The Minkowski sum of convex polygons is a convex polygon.

The Minkowski sum of two line segments is a parallelogram.
The Minkowski sum of two triangles is a hexagon (usually not tangential).
Any convex polygon is the Minkowski sum of triangles and line segments. A
reference for this (which I have yet to check) is page 177 of
I. M. Yaglom, V. G. Boltyanskii, (1961) Convex Figures, New York: Holt,
Rinehart and Winston.
This leads on to the following.

Questions. (i) Is any convex hexagon the Minkowski sum of 2 triangles?
(ii) Is, for $n\geq{6}$, any convex $n$-gon with origin at the centroid the
Minkowski sum of a small number (perhaps just 2) of tangential $m$-gons (with
$m\leq{n/2}$), now (unlike everywhere else in this document) all with centroid
at the origin?
A positive answer to a question like this might suggest, from properties
established for all tangential $m$-gons, corresponding properties for convex
$n$-gons. And if this were to be the case, one can imagine establishing some
domain-functional property for tangential $n$-gons and then getting something
from the concavity of the domain-functional under Minkowski sum.

###  12.9 Rearrangements???

For the purely geometric functionals $A$, $L$, $i_{2}$, $i_{4}$, $Q_{0-}$,
$d_{O}$ studied in this document rearrangements might not be needed. For
functionals, like conformal inradius, transfinite diameter, torsional
rigidity, fundamental frequency – functionals involving integrals of gradient
squared, etc. – rearrangements are an appropriate tool. [69] used Steiner
symmetrization to establish, for triangles and quadrilaterals, that, at given
area, the regular $n$-gon optimizes. These are equilateral triangle and square
respectively. However, Steiner symmtrizing polygons with more vertices
typically increases the number of vertices. [83, 82, 84] manage to use some
sort of rearrangement preserving the number of vertices of a convex $n$-gon.
Dissymetrization?
ToDo. Try to understand this. See also [4]

Polarizations seem to be building blocks for the rearrangements with which I
am more familiar, namely those used in [69]. Polarizations, alone, are not
likely to be a tool for rearranging tangential polygons. The following guesses
and questions began with drawing sketches.

  * •

The polarization of a triangle about an angle bisector just reflects the
triangle.

  * •

Can anything beyond the incircle staying fixed be said about the polarization
of a triangle about any line through the incentre? Similar question for any
tangential polygon $\Omega$ about any line through the incentre? Non-convex
circumgons?

[82]
(i) presents results of the kind that regular $n$-gons optimize over all
$n$-gons with the same area;
(ii) that tangential polygons are used (see Part III §22).
A process called ‘dissymetrization’ gets used. This, and polarization, get a
mention in [84].

###  12.10 Spaces of polygons?

See [28].

##  13 Construction of tangential polygons

With just a few exceptions (on 1-cap, etc.) to date our computations have been
for tangential $n$-gons.

Our first method provided data for actually drawing up the polygon and
required an initial specification of the inradius. After this prescribe $n$
(which in our computations so far just $n\leq{6}$). Then choose $n$ increasing
values of $\theta_{k}$ in $(-\pi,\pi)$ with the maximum difference of
consecutive $\theta_{k}$ less than $\pi$. This yields $\exp(\theta_{k})$ on
the unit circle as tangent points. (The restriction on the separation of the
$\theta_{k}$ is in order that a convex polygon is constructed.) From each pair
of consecutive tangent points, find the point of intersection of the tangent
lines. These give the vertices of the tangential polygon and there are
standard formulae for perimeter $L$, area $A$, in terms of the coordinates but
it is easier to note that with the tangent lengths one can find the $T_{k}$
and use these to find $L$, $i_{2}$, $i_{4}$, etc.

If one doesn’t need to draw the polygon and is interested in functionals that
stay constant under change of scale, e.g. $L/\rho$, one can use the $T_{k}$ as
defined in Part IIa equation (6.8). The tangent lengths are given by
$\eta_{k}=\rho\,T_{k}$. Simple formulae to find the other functionals $L$,
$i_{2}$, $i_{4}$ are given Part IIa §6.1, and another $d_{O}$ in §19.

We start from result given at the beginning of §12:
Existence Theorem.Given a convex $n$-gon, and hence its sequence of angles,
all less than $\pi$, then one can define a tangential $n$-gon whose incentre
is at the origin, with the same sequence of angles.
Clearly once one has one, one can rescale by any factor preserving the
properties.

Corollary. Given a set $S$ of $n$-numbers $0<\alpha_{k}<\pi$ summing to
$(n-2)\pi$, then the different tangential $n$-gons arising from the different
permutations of $S$ all have the same values for

$\frac{L}{\rho},\ \ \frac{i_{2}}{\rho^{3}},\ \ \frac{i_{4}}{\rho^{5}},\ \
\frac{d_{O}}{\rho}.$

In any event, for many calculations later in this part, one can start directly
with the tangent lengths, or with the $\alpha_{k}$ or with the $T_{k}$: it is
not always necessary to calculate the vertex coordinates.

In some contexts moving between tangential $n$-gons by permuting angles may
lose properties. In particular, permuting angles of a bicentric $n$-gon will,
in general result in a tangential $n$-gon which is not bicentric. (The
simplest example would be any bicentric quadrilateral with 4 different angles.
Permuting the angles must lose the property that the sum of opposite angles is
$\pi$.)

While $d_{O}$, the distance from the incentre origin to a vertex, doesn’t need
the vertex coordinates and and is invariant, at fixed $\rho$ under
permutations of the angles, this may not be the case for the circumradius $R$.

##  14 Duality

Denote the inner product for plane vectors with a dot. Define polar-
reciprocation $\cal P$ of a point $x$ by

${\cal P}(x)=\\{z\ |\ z\cdot{x}=1\\}.$

$\cal P$ takes a point to a line. (This differs from the most common
definition of polar in convex geometry in which one has $z\cdot{x}\leq{1}$ so
points map to half-planes.) Next continue the definition. Let $D$ be a set in
the plane. Define ${\cal P}(D)$ by

${\cal P}(D)=\\{z\ |\ z\cdot{x}=1\ \forall x\in{D}\\}.$

$\cal P$ takes the unit circle to itself. $\cal P$ takes lines (not through
the origin) to points: in particular $\cal P$ takes a line tangent to the unit
circle to its point of tangency with the unit circle.

See
`https://en.wikipedia.org/wiki/Pole_and_polar`
`https://en.wikipedia.org/wiki/Dual_polygon`

Thus the boundary lines of a tangential polygon map to the vertices of a
cyclic polygon and vice-versa.

‘Vertex-side’ duality, adapted from wikipedia

As an example of the side-angle duality of polygons we compare properties of
the cyclic and tangential polygons, especially quadrilaterals.

Cyclic $n$-gon |  Tangential $n$-gon
---|---
Circumscribed circle | Inscribed circle
Perpendicular bisectors of the sides are | Angle bisectors are
concurrent at the circumcentre | concurrent at the incentre
$n$ even: The sums of the two pairs |  $n$ even: The sums of the two pairs
of opposite/alternate angles | of opposite/alternate sides
are equal | are equal

For a cyclic $2m$-gon the sum of the alternate angles is $(m-1)\pi$. We remark
that for quadrilaterals, $m=4$ $n=2$ the converse is true but it is false for
$n\geq{3}$. The same is true for tangential $2m$-gons: for $m\geq{3}$ being
‘2-special’ is necessary but not sufficient for a $2m$-gon to be tangential.

$n=6$, $m=3$. Brianchon’s theorem states that the three main diagonals of a
tangential hexagon are concurrent.
The polar reciprocal and projective dual of the conics version of Brianchon’s
theorem give Pascal’s theorem.

Duality is evident again when comparing an isosceles trapezoid to a kite.

Isosceles trapezoid | Kite
---|---
Two pairs of equal adjacent angles | Two pairs of equal adjacent sides
One pair of equal opposite sides | One pair of equal opposite angles
An axis of symmetry through | An axis of symmetry through
one pair of opposite sides | one pair of opposite angles
Circumscribed circle | Inscribed circle

Let $P(n)$ be an ordered list of $n$ points $\zeta_{k}$ on a circle,
$P(n+1)=P(n)\cup\\{\zeta_{n+1}\\}$ with $\zeta_{n+1}$ after $\zeta_{n}$ and
before $\zeta_{1}$.
Let $TP(P(n))$ be the tangential $n$-gon with the points of $P(n)$ its tangent
points.
Let $CP(P(n))$ be the cyclic $n$-gon with the points of $P(n)$ its vertices.
The first entry in the table below indicates how areas change on introducing
the additional point on the circle.

$|TP(P(n+1))|\leq|TP(P(n))|$ | $|CP(P(n+1))|\geq|CP(P(n))|$
---|---
A tangential $2m$-gon has all sides equal iff | A cyclic $2m$-gon has all angles equal iff
the alternate angles are equal. | the two sets of alternate sides are equal.
Equilateral tangential for $n=2m$ even | Equiangular cyclic for $n=2m$ even
Opposite angles equal if $n/2$ is even | Opposite sides equal if $n/2$ is even

See [90].

###  14.1 Tangential and cyclic polygons, continued

The set of all convex sets is a lattice under operations of intersection,
$\Omega_{1}\cap\Omega_{2}$, and convex-hull of union, ${\rm
conv}(\Omega_{1}\cup\Omega_{2})$.
The set of tangential polygons with incentre at the origin and given inradius
$\rho$ is closed under intersection.
The set of cyclic polygons with circumcentre at the origin and given
circumradius $R_{V}$ is closed under convex-hull-union. (A cyclic polygon is
the convex hull of its extreme points all of which lie on the circle radius
$R_{V}$.)
In the case of $n$-gons the numbers of vertices can increase.

Let the coordinates of the vertices of a convex $n$-gon be $(x_{k},y_{k})$
with the vertices traversed in order (and vertex 1 identified with vertex
$n+1$).
The polygon is cyclic with circumcentre O and circumradius 1 if the distance
of every vertex from O is 1:

$x_{k}^{2}+y_{k}^{2}=1\qquad\forall k.$

The polygon is tangential with incentre O and inradius 1 if the distance of
every side from O is 1:

$(x_{k+1}-x_{k})^{2}+(y_{k+1}-y_{k})^{2}=(x_{k+1}y_{k}-y_{k+1}x_{k})^{2}\qquad\forall
k.$

The tangency point on each line, the closest point to O, is

$x_{t}=\frac{y_{k+1}-y_{k}}{x_{k}\,y_{k+1}-y_{k}\,x_{k+1}},\qquad
y_{t}=\frac{x_{k+1}-x_{k}}{x_{k}\,y_{k+1}-y_{k}\,x_{k+1}}.$

In establishing, by Steiner symmetrisation, isoperimetric properties of
$n$-gons, for $n=3$ and $n=4$, sequences of polygons which alternate between
tangential and cyclic occur. Here is a quote from [69]:

> Of all quadrilaterals with a given $A$, the square has the smallest $L$,
> $I_{c}$ (polar moment of inertia about the centroid), $\overline{r}$, …but
> the largest $\dot{r}$ and $Q_{0}$. ……it is sufficient to indicate a sequence
> of symmetrizations which transform, ultimately, a given quadrilateral into a
> square. Symmetrizing a given quadrilateral with respect to a perpendicular
> to one of ·its diagonals, we change it into a quadrilateral having a
> diagonal as axis of symmetry. Symmetrizing this new quadrilateral with
> respect to a perpendicular to its axis of symmetry, we change it into a
> rhombus. Symmetrizing the rhombus with respect to a perpendicular to one of
> its sides, we change it into a rectangle. Symmetrizing the rectangle with
> respect to a perpendicular to one of its diagonals, we obtain another
> rhombus. Repeating the last two steps in succession, we obtain an infinite
> sequence in which rhombi alternate with rectangles.

Rhombi are tangential polygons (with equal sides):
rectangles are cyclic (with equal angles).

Steiner symmetrization is not applicable to showing that regular $n-gons$
optimize when $n\geq{5}$. If one (initially at least) focuses on geometric
quantities like

$\frac{A}{L^{2}},\ \frac{i_{2}}{L^{3}},\ \frac{i_{4}}{L^{4}},\qquad{\rm and\
}\ \frac{Q_{0-}}{A^{2}},$

it may be possible to devise other transformations between $n$-gons
which alternate between tangential and cyclic,
which change functionals like those immediately above monotonically and
which converge to the regular $n$-gon.

##  15 Miscellaneous properties of tangential polygons

###  15.1 Circulant matrices and tangential $n$-gons

This subsection treats questions like the following:
Given a set of $n$ of positive side lengths $(s_{j})$ how can we recognize if
there could be a tangential polygon with these side lengths?

We begin with a connection between tangential polygons and circulant matrices
presented near the beginning of the wikipedia page on tangential polygons.

Let $P$ and $M=I+P$ be the $n\times{n}$ circulant matrices as follows. $P$ is
the $n\times{n}$ cyclic permutation:

$P=\begin{pmatrix}0&1&0&\cdots&0\\\ 0&0&1&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\ 1&0&0&\cdots&0\end{pmatrix}$

The matrix $\frac{1}{2}M$ is doubly stochastic.

The wikipedia page states:

> There exists a tangential polygon of $n$ sequential sides
> $s_{1},\ldots,s_{n}$ if and only if the system of equations
>
> $M{\mathbf{\eta}}={\mathbf{s}},$ (15.1)
>
> has a solution $(\eta_{1},\dots,\eta_{n})$ in positive reals. If such a
> solution exists, then $(\eta_{1},\dots,\eta_{n})$ are the tangent lengths of
> the polygon (the lengths from the vertices to the points where the incircle
> is tangent to the sides).

Once one has the $\eta$ and $\rho$ the angles $\alpha_{k}$ are determined from

$\eta_{k}(=\eta_{k-})=\rho T_{k}\qquad{\rm where\ \
}T_{k}=\frac{1}{\tan(\frac{\alpha_{k}}{2})}.$

See PartIIa, equation (6.8). That the sum of the $\alpha_{k}$ is $(n-2)\pi$
leads to one further equation which we record, as an aside, in the next
subsubsection.

####  15.1.1 Some relations between the $T_{k}$

As before, consider $n$-gons and denote the angle at vertex $k$ by
$\alpha_{k}$, with $k$ increasing as one goes around the convex $n$-gon in
counterclockwise direction. The sum over all the $\alpha_{k}$ is $(n-2)\pi$.

Fix the inradius $\rho$ as 1. Suppose the points of tangency of the tangential
$n$-gon are $\zeta_{j}=\exp(i\phi_{j})$. Again $j$ increases as one goes
around the convex $n$-gon in counterclockwise direction. The angle at O formed
by the lines $O\zeta_{j}$ and $O\zeta_{j+1}$ is $\phi_{j+1}-\phi_{j}$.

Then, with

$T_{k}=\frac{1}{\tan\frac{\alpha_{k}}{2}}=\tan(\frac{\pi-\alpha_{k}}{2}),$
(15.2)

$T_{k}>0$ since the polygon is convex. Since we know the sum of
$\alpha_{k}/2$:

$\sum_{k=1}^{n}\frac{\alpha_{k}}{2}=\sum_{k=1}^{n}{\rm
arccot}(T_{k})=(n-2)\frac{\pi}{2}.$ (15.3)

Some, but not all, the information in this can be expressed in equations
involving just rational functions of the $T_{k}$, i.e. without the
transcendental arccot function. We denote the elementary symmetric polynomial
of degree $k$ by

${\rm SymmetricPolynomial}(k,\ldots),$

and define

$e_{k}={\rm
SymmetricPolynomial}(k,[\frac{1}{T_{1}},\frac{1}{T_{2}},\ldots,\frac{1}{T_{n}}]).$

For tangential $n$-gons, we first treat $n$ odd, then $n$ even.
When $n$ is odd $\cos((n-2)\pi/2)=0$ so the cosine of the sum of all the
$\alpha_{k}/2$ is $0$ we have

$e_{0}-e_{2}+e_{4}-e_{6}\dots=0.$ (15.4)

When $n$ is even $\sin((n-2)\pi/2)=0$ so the sine of the sum of all the
$\alpha_{k}/2$ is $0$ we have

$e_{1}-e_{3}+e_{5}-e_{7}\dots=0.$ (15.5)

We will need the formulae for perimeter, i.e. $L=i_{0}$, for $i_{2}$ and for
$i_{4}$ as given in Part IIa §6.1 namely equations (6.9), (6.10) and (6.11).

####  15.1.2 Examples at $n=3$ or $4$

For a triangle
$A={\sqrt{\frac{L}{2}(\frac{L}{2}-a)(\frac{L}{2}-b)(\frac{L}{2}-c)}}$ and
since $L=2\sum\eta_{k}$

$None$

For a tangential quadrilateral wikipedia gives

$None$

Figure 4: From wikipedia. Brianchon’s Theorem. Diagonals of a tangential
hexagon are concurrent.

####  15.1.3 Non-negative solution for $\mathbf{\eta}$ given $\mathbf{s}$?

Return now to the linear equations (15.1). Denote the dependence of $M$ on $n$
by writing $M(n)$. There is very different behaviour when $n$ is odd than when
$n$ is even as, for example,

${\rm det}(M(n))=1-(-1)^{n}.$

Thus $M(n)$ is invertible when $n$ is odd, but not when $n$ is even. There
being many questions I have been, as yet, unable to answer, and as properties
of $M(n)$ might ultimately be useful, I have collected several properties of
$M(n)$ here, but not found uses for some of them yet. Let ${\mathbf{e}}$ be
the vector all of whose entries are 1. Then, for all $n$,

$M(n)\,{\mathbf{e}}=2\,{\mathbf{e}}.$

Geometrically this corresponds to a regular $n$-gon with side 2 and tangent
lengths 1. The eigenvalues of $P$ are the roots of unity, and the eigenvalues
of $M$ require us just add 1 to these. The eigenvectors are, of course, the
same.

${\rm CharacteristicPolynomial}(M(n),\lambda)=-(-1)^{n}(1-(\lambda-1)^{n}).$

The transpose $M^{T}(n)$ has exactly the same properties as described for
$M(n)$ in the preceding paragraph. If one ignores the nonnegativity
requirement, clearly there is, for any rhs $\mathbf{s}$ a unique ‘solution’ to
the linear equations when $n$ is odd. When $n$ is even, there is only a
‘solution’ when the rhs is orthogonal to the nullspace of $M^{T}(n)$, i.e only
when

$\sum_{k\ {\rm odd}}s_{k}=\sum_{k\ {\rm even}}s_{k},$ (15.6)

and, also, ‘solutions’ are not unique.

$M(n)$ is normal, commutes with its transpose. As $M\,M^{T}$ is symmetric and
Toeplitz it is centrosymmetric.

A persymmetric matrix is a square matrix which is symmetric with respect to
the northeast-to-southwest diagonal. $M(n)$ is persymmetric and, as such,
satisfies

$M(n)\,J=J\,M^{T}(n),\qquad{\rm where\ }J\ {\rm is\ the\ exchange\ matrix}.$

$J$ is the matrix with 1s on its northeast-to-southwest diagonal and 0s
elsewhere.

Define next

$M_{i}(n)=(I(n)+\sum_{k=1}^{n-1}(-1)^{k}P(n)^{k})/2.$

We have

$M(n)\,M_{i}(n)=\frac{1-(-1)^{n}}{2}I(n).$

For $n$ odd $M_{i}(n)$ is the inverse of $M(n)$. Also (via Cayley-Hamilton
Theorem)

$\sum_{k=1}^{n}(-1)^{k}{n\choose k}M(n)^{k}=-(1-(-1)^{n})\,I(n).$

The obvious next question is ‘what conditions on the sides ensure there is a
nonnegative solution for $\mathbf{\eta}$’?

Farkas Lemma. Exactly one of the following two assertions is true:
1.There exists an ${\mathbf{\eta}}\in{\mathbb{R}^{n}}$ such that
$M{\mathbf{\eta}}={\mathbf{s}}$ and ${\mathbf{\eta}}\geq 0$.
2\. There exists a $\mathbf{y}\in\mathbb{R}^{n}$ such that
$\mathbf{M}^{\mathsf{T}}\mathbf{y}\geq 0$ and
$\mathbf{s}^{\mathsf{T}}\mathbf{y}<0$.
We have yet to devise a good use for this lemma but suspect it will be
relevant to conditions for general $n\geq{3}$.

Here we do not attempt to find the conditions for general $n$. We will attempt
to find necessary and sufficient conditions on $\mathbf{s}$ for the existence
of nonnegative $\mathbf{\eta}$, separately, for each of $n=3$, 4, 5 and 6.

Triangles. When $n=3$,

$M(3)^{-1}=\frac{1}{2}\left(\begin{array}[]{ccc}1&-1&1\\\ 1&1&-1\\\ -1&1&1\\\
\end{array}\right),$

so one has nonnegative solutions for $\eta$ iff the nonnegative $s$ are such
that the sum of any two is greater than (or equal to) the remaining side’s
length. This accords with the fact that any triangle is tangential (indeed
bicentric).

Tangential pentagons. When $n=5$,

$M(5)^{-1}=\frac{1}{2}\left(\begin{array}[]{ccccc}1&-1&1&-1&1\\\
1&1&-1&1&-1\\\ -1&1&1&-1&1\\\ 1&-1&1&1&-1\\\ -1&1&-1&1&1\\\
\end{array}\right).$

Consider triples of sides in which just one pair of sides are adjacent. There
are five such triples. There are nonnegative solutions for $\eta$ iff the
nonnegative $\mathbf{s}$ are such that for any of these five triples the sum
of those in the triple is greater than (or equal to) the sum of the remaining
two sides’ lengths.

Let $n=2m$ for $m\geq{2}$. Define

${\mathbf{nv}}=\left(\ (-1)^{k}\ \right),$

which is a basis for the nullspace of $M(n)$ (and also of $M^{T}(n)$). Suppose
$\mathbf{s}$ satisfies equation (15.6) so ‘solutions’, albeit without the
nonnegativity condition imposed, exist. These solutions are

${\mathbf{\eta}}_{\rm gen}={\rm
PseudoInverse}(M(n))\,{\mathbf{s}}+c\,{\mathbf{nv}},$ (15.7)

but it remains to investigate the restrictions on $\mathbf{s}$ and $c$ needed
so that amongst the ${\mathbf{\eta}}_{\rm gen}$ there is at least one with
${\mathbf{\eta}}\geq{0}$. In this document we will attempt this only for $n=4$
and $n=6$. Before that, however, we consider $n=2m$ in general. By standard
properties

$M(n)\,{\rm PseudoInverse}(M(n))\,M(n)=M(n).$

Using the fact that when $n=2m$ the matrix $M(n)$ has the simple block
structure

$M(n)=\begin{pmatrix}U&L\\\ L&U\end{pmatrix},$

it is easy to find $A$ and $B$ so that

${\rm PseudoInverse}(M(n)=\begin{pmatrix}A&B\\\ B&A\end{pmatrix}.$

The matrix $L$ has just one 1 in the bottom left corner and $U=M(m)-L$. We
have

$\displaystyle A$ $\displaystyle=$ $\displaystyle\frac{1}{2}({\rm
PseudoInverse}(U+L)+{\rm PseudoInverse}(U-L)),$ $\displaystyle B$
$\displaystyle=$ $\displaystyle\frac{1}{2}({\rm PseudoInverse}(U+L)-{\rm
PseudoInverse}(U-L)).$

We also have $L\,U\,L$ is the zero matrix, and $L\,U^{-1}\,L=-L$.

Tangential quadrilaterals. It is already known that no further restriction
beyond

$s_{1}+s_{3}=\frac{L}{2}=s_{2}+s_{4}$

is needed to ensure that the quadrilateral is tangential. So it remains just
an exercise to show that the system of equations

$M(4){\mathbf{\eta}}=\begin{pmatrix}s_{1}\\\ s_{2}\\\ \frac{L}{2}-s_{1}\\\
\frac{L}{2}-s_{2}\end{pmatrix}$

has, for all $0<s_{1}<L/2$ and $0<s_{2}<L/2$ a positive $\eta$ solution. When
$n=4$,

${\rm PseudoInverse}(M(4))=\frac{1}{8}\left(\begin{array}[]{cccc}3&-1&-1&3\\\
3&3&-1&-1\\\ -1&3&3&-1\\\ -1&-1&3&3\\\ \end{array}\right).$

There is no loss of generality in setting $L=2$ and considering
$1/2\leq{s_{1}}<1$ and $1/2\leq{s_{2}}<s_{1}$. If we try the formula (15.7) we
are led to consider the function $\phi$

$\phi(s_{1},s_{2},c)={\rm
min}(1+2s_{1}-2s_{2}-c,-1+2s_{1}+2s_{2}+c,1-2s_{1}+2s_{2}-c,3-2s_{1}-2s_{2}+c),$

over the triangle in $(s_{1},s_{2})$ space defined in the preceding sentence.
Considering the final entry in the min defining $\phi$, we have
$\phi(s,s,0)<0$ for $3/4<s<1$, so we need to choose $c$ (which can depend on
$\mathbf{s}$) appropriately. We find, over the triangle in
$(s_{1},s_{2}$-space $\phi(s_{1},s_{2},2s_{1}+2s_{2}-3)=0$, i.e. the last
entry is zero but the other 3 entries are nonnegative. Except for a positive
multiple, the other 3 entries (first 3) are

$1-s_{2},-1+s_{1}+s_{2},1-s_{1}.$

Tangential hexagons. Unlike the situation with $n=4$ extra conditions on the
sides are needed. The corresponding problem for cyclic hexagons is mentioned
in [91] (and the same author has other papers involving tangential and cyclic
hexagons, [88], [90]).

After writing the above, I found [15] gives the following.

Theorem.There will be a tangential hexagon with given side lengths
$s_{1},s_{2},...,s_{6}$ if and only if the equality

$s_{1}+s_{3}+s_{5}=s_{2}+s_{4}+s_{6},$

and the following nine inequalities are satisfied:

$s_{1}>0,s_{2}>0,...,s_{6}>0,$ $\displaystyle s_{1}-s_{2}+s_{3}$
$\displaystyle>$ $\displaystyle 0,$ $\displaystyle s_{3}-s_{4}+s_{5}$
$\displaystyle>$ $\displaystyle 0,$ $\displaystyle s_{5}-s_{6}+s_{1}$
$\displaystyle>$ $\displaystyle 0.$

In words, the length of any side is less than the sum of the lengths of the
adjacent sides.

[15] also gives the conditions on $\mathbf{s}$ for an octagon to be
tangential. When $n=6$,

${\rm
PseudoInverse}(M(6))=\frac{1}{12}\left(\begin{array}[]{cccccc}5&-3&1&1&-3&5\\\
5&5&-3&1&1&-3\\\ -3&5&5&-3&1&1\\\ 1&-3&5&5&-3&1\\\ 1&1&-3&5&5&-3\\\
-3&1&1&-3&5&5\\\ \end{array}\right).$

(On looking at the corresponding outputs for large $n=2m$ one finds that $2n$
times ${\rm PseudoInverse}(M(n))$ has entries $\pm$ odd integers less than
$n$. And, as noted before, there is a block matrix structure too.)

ToDo. Check out that the conditions on $\mathbf{s}$ given in [15] are
necessary and sufficient to ensure that $\eta$ is nonnegative.

###  15.2 Geometric isoperimetric inequalities

Lets begin with the first, historic, instance of an isoperimetric inequality
in which regular $n$-gon optimizes over all $n$-gons. The following argument
is from geometers in ancient Greece, perhaps around 200BC.

L-A Isoperimetric Result. Amongst convex $n$-gons with given perimeter, that
which has the largest area is the regular $n$-gon.

Start with a convex $n$-gon, $\Omega_{0}$. Suppose $P_{i-1}$, $P_{i}$,
$P_{i+1}$ are consecutive vertices. (i) show that among all isoperimetric
triangles with the same base the isosceles triangles has maximum area. Thus by
changing $\Omega_{0}$ moving point $P_{i}$ to $P_{i}^{\prime}$ with $P_{i-1}$,
$P_{i}^{\prime}$, $P_{i+1}$ isosceles, the area of the changed polygon is
increased. Apply this process to all triples of consecutive vertices. By
iteration, one finds that the optimal polygon must be equilateral: call it
$\Omega_{1}$.
(ii) show that if the polygon $\Omega_{1}$ is not equiangular, its area may be
increased by redistributing perimeter from a pointy to a blunt angle until the
two angles are the same.

An alternative to (ii) is to use that Among all $n$-gons with given side
lengths, the cyclic $n$-gon has the largest area. And a cyclic $n$-gon with
equal sides is regular.

(For this and related isoperimetric inequalities see [3]. See also [90]
concerning equiangular and equilateral considerations.)

By a $\cal D$ we shall mean some domain functionals, and we are interested in
pairs of these for which one has a result of the form
For tangential $n$-gons with fixed ${\cal D}_{1}$ the regular $n$-gon
$<$maximizes$|$minimizes$>$ ${\cal D}_{2}$
Table 4 below presents some:

${\cal D}_{1}$ |  | ${\cal D}_{2}$ | Remark
---|---|---|---
area | min | perimeter |
area | max | inradius | $A=\rho L/2$
inradius | min | perimeter | Jensen inquality
inradius | min | area | ”
inradius | min | $i_{2}=\frac{16\Sigma_{\infty}}{\rho}$ | ”
inradius | min | $i_{4}$ | ”
Table 4: Tangential $n$-gons

The Jensen inequality/convexity concerns convex functions $\phi$ on some
interval $I$, and that

${\rm with}\ x_{i}\in{I}\ {\rm and}\ \sum a_{i}=1\ {\rm with}a_{i}\geq{0}\ ,\
\phi(\sum a_{i}x_{i})\leq\sum a_{i}\phi(x_{i}).$

The inequality is reversed for concave $\phi$. By using the formula for the
length for a $n$-gon given at equation (6.9) and the convexity of
$\cot(\cdot/2)$ on $(0,\pi)$ we have the first entry of the following.

  * •

We use $\cot(\alpha/2)$ convex for $\alpha\in(0,\pi)$ and (6.9). We have

$\frac{L}{2\rho
n}=\sum\frac{1}{n}\cot(\frac{\alpha_{k}}{2})\geq\cot\left(\sum\frac{1}{n}\frac{\alpha_{k}}{2}\right)=\cot(\frac{(n-2)\pi}{2n})\frac{L_{n}}{2\rho
n},$

where $L_{n}$ is the perimeter of the regular $n$-gon with inradius $\rho$.
This establishes the entry in the table corresponding to fixed $\rho$,
mimimizing the perimeter (and since $A=\rho\,L/2$ also minimiing $A$).

  * •

Using that

$\cot(\frac{\alpha}{2})+\frac{\cot(\frac{\alpha}{2})^{3}}{3}\ \ {\rm convex\
for\ }\alpha\in(0,\pi),$

and the formula (6.10) the Jensen inequality approach above gives that, at
fixed $\rho$, $i_{2}$ is minimized by the regular polygon with the same number
of sides.

  * •

Starting from formula (6.11), the result on $i_{4}$ follows in the same way.

For cyclic $n$-gons with fixed ${\cal D}_{1}$ the regular $n$-gon
$<$maximizes$|$minimizes$>$ ${\cal D}_{2}$
Table 5 below presents some:

${\cal D}_{1}$ |  | ${\cal D}_{2}$ | Remark
---|---|---|---
area | min | perimeter |
area | max | inradius |
circumradius | max | perimeter | Jensen inquality
circumradius | max | area | ”
Table 5: Cyclic polygons

There is a huge literature even restricting to convex sets.
`https://math.stackexchange.com/questions/749528/isoperimetric-inequality-
isodiametric-inequality-hyperplane-conjecture-what`

Also the ‘stability’ of the isoperimetric inequalities is studied in many
different ways sometimes involving ‘isoperimetric deficit’, sometimes
‘Fraenkel asymmetry’.

If $L$ is the perimeter of a convex polygon $\Omega$, $A$ its area, $\rho$ the
inradius, and $s$ the length of any chord through the centre of a largest
inscribed circle, then

$L^{2}-4\pi A\geq\frac{\pi^{2}}{4}(s-2\rho)^{2}.$

This is a sharpened isoperimetric inequality for convex polygons.
H. Hadwiger, Comment. Math. Helv. 16 (1944),305-309.
Tangential polygons can be regarded as limits of tangential $n$-gons, so the
inequality applies to them and is

$L(L-2\pi\rho)\geq\frac{\pi^{2}}{4}(s-2\rho)^{2}.$

In notation as in the list at the beginning of §19,

$s\geq d_{O}+\rho\qquad{\rm so}\ \
L(L-2\pi\rho)\geq\frac{\pi^{2}}{4}(d_{O}-\rho)^{2}.$

With, as in §19.5, $x=L/d_{O}$, $y=\rho/d_{O}$,

$x\,(x-2\pi y)\geq\frac{\pi^{2}}{4}(1-y)^{2},$

or

$x=\frac{1}{2}\pi\left(\sqrt{5y^{2}-2y+1}+2y\right).$

This gives a curve bit left of the upper left 1-cap curve of Figure 12.

There is a considerable literature concerning moments of inertia about the
centroid. (There are, of course, situations involving appropriate symmetries
when the centroid and incentre will coincide.)
Results. The equilateral triangle minimizes the moment of inertia, among all
convex curves with given perimeter.
References include [40], [85], [64]. See also §20.

###  15.3 Further geometric items

For every convex domain

$\pi\rho+\frac{|\Omega|}{\rho}\leq|\partial\Omega|\leq
2\,\frac{|\Omega|}{\rho}.$

See [20] equation (8). For tangential polygons the right hand side is an
equality and the left-hand side is just $L\geq{2}\pi\rho$.
There may be some use for Bonnesen inequalities, see [66], and stronger forms
for tangential polygons.

Repeat here, for the third time(!), the existence statement of §12 and of §13.
Let there be given a convex polygon, and hence its sequence of angles, all
less than $\pi$. Then one can define a tangential polygon with the same
sequence of angles and the same area.

A tangential polygon has a larger area than any other convex polygon with the
same perimeter and the same interior angles in the same sequence. (See [2, 54,
93].)
Amongst all convex polygons with the same area and with the same interior
angles in the same sequence
(i) those which have the smallest perimeter are tangential polygons, and
(ii) those which have the largest inradius are tangential polygons.

Amongst all tangential quadrilaterals with a given sequence of side lengths,
that which maximizes the area is bicentric.
One starting point is the more general result.
Amongst all quadrilaterals with given side lengths, that which has maximum
area is cyclic.
(Proof: Use Bretschneider’s formula.)
(A very easy special case is that the area of a kite with given sides is
maximized by the right kite.)

There may be generalization to $2m$-gons, in particular hexagons.
`https://mathworld.wolfram.com/CyclicHexagon.html`
gives area in terms of sides.

Theorem. For any quadrilateral with given edge lengths, there is a cyclic
quadrilateral with the same edge lengths.

Theorem. The cyclic quadrilateral has the largest area of all quadrilaterals
with sides of the same length.

###  15.4 Cheeger constant

For tangential polygons $\Omega$, the Cheeger constant is

$h_{\Omega}=\frac{|\partial\Omega|+\sqrt{4\pi|\Omega|}}{|2\Omega|}.$

Clearly, at fixed area, since perimeter is minimized by the regular $n$-gon,
the regular $n$-gon minimizes $h_{\Omega}$ over tangential $n$-gons with given
area. Much more has been established.

Among all simple polygons with a given area and at most $n$ sides, the regular
$n$-gon minimizes the Cheeger constant. (See [16].)

If $\Omega$ is a convex polygon, we denote $\Omega_{*}$ the (unique up to
rigid motions) circumscribed polygon which has the same area as $\Omega$ and
whose angles are the same as those of $\Omega$, then

$h(\Omega)\geq h(\Omega_{*}),$

with equality if and only if $\Omega=\Omega_{*}$ (up to rigid motions).

##  16 Triangles

Given the area $A$ and the angles $\alpha$, $\beta$ and
$\gamma=\pi-\alpha-\beta$ of the triangle, one can determine the inradius and
thence, if needed, the sides. We have

$A=\rho^{2}\left(\cot\frac{\alpha}{2}+\cot\frac{\beta}{2}+\cot\frac{\gamma}{2}\right).$

The semiperimeter $s=(a+b+c)/2$ is found from

$s=\frac{A}{\rho}=\rho\,\left(\cot\frac{\alpha}{2}+\cot\frac{\beta}{2}+\cot\frac{\gamma}{2}\right)=\sqrt{A\,\,\left(\cot\frac{\alpha}{2}+\cot\frac{\beta}{2}+\cot\frac{\gamma}{2}\right)}.$

Now, defining $f$ from

$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}=\frac{1}{f},$

we have

$s=\frac{a+b+c}{2}=\frac{f}{2}\,\left(\sin(\alpha)+\sin(\beta)+\sin(\gamma)\right),$

which determines $f$ and hence all the sides.

Using

$T_{A}=\frac{1}{\tan(\frac{\alpha}{2})},\quad
T_{B}=\frac{1}{\tan(\frac{\beta}{2})},\quad
T_{C}=\frac{T_{A}+T_{B}}{T_{A}T_{B}-1},$

$i_{2}$ and $i_{4}$ can be found from equations (6.10) and (6.11). From these
$\Sigma_{\infty}$ and $\Sigma_{1}$ can be found: see Part III §25

With apologies for leaving the write-up in code, all the reasonable results
are readily proved for triangles. The code also produces an example of a
Blaschke-Santalo diagram, shown in Figure 5, of the kind we will see later for
other tangential polygons.

    ( * An isoperimetric result
    Amongst triangles with a given inradius, taken as 1,
    that which has the smallest perimeter is the equilateral triangle.
    The allowed values of TA>0 and TB>0 are those for which TA*TB>1 *)
    TCfn[TA_,TB_]:= (TA+TB)/(TA*TB-1);
    Lfn[TA_,TB_]:= TA+TB+TCfn[TA,TB];
    (* Lfn ia half the length *)
    (* The hessian is positive definite, positive diagonal entries and Det>0
    on using TA*TB>1, etc. *)
    hess = Map[Simplify, D[tmp, {{TA, TB}, 2}]];
    Factor[Det[hess]]
    (* Find minimum by checking where gradient is 0  *)
    g = Map[Simplify, Grad[Lfn[TA, TB], {TA, TB}]]
    Solve[{g[[1]] == 0, g[[2]] == 0}, {TA, TB}]
    (* gives (TA,TB) = (Sqrt[3],Sqrt[3]) which is equilateral triangle *)

    (* Define also *)
    dOfn[TA_,TB_]:= Max[{TA,TB,TCfn[TA,TB]}];
    (* One can also show that at given inradius, the triangle which
    minimizes the distances from incentre to vertices is equilateral. *)

    (* One can also imagine fixing not just the inradius but also dO
    and with these TWO constraints finding
    (i) the shape with the maximum perimeter,
    (ii) the shape with the minimum perimeter.
    And find that they come out to be the obvious different sorts of isosceles triangles.
    *)

    (* Think of A as apex of triangle
    Expect to see different behaviour for TA large - small apex angle
    to what one gets when TA is small - big apex angle.
    Let B the angle I will vary be <= Pi/2 so TB >=1.
    Actually TB also gets restricted to be TB >= Max[1,1/TA]
    This restiction causes TC>0 *)

    (* fix TA, vary TB
     LAfn is a convex function and its first derivative is zero when TB=TC,
     i.e. triangle is isosceles
     So, at fixed rho, the perimeter of a triangle with one angle fixed is
    minimized when the triangle is isosceles having TB=TC= (1+Sqrt[1+TA^2])/TA
    BCequal = Factor[TCfn[TA, TB] - TB]
    Solve[BCequal == 0, TB]
    Factor[D[Lfn[TA, TB], TB]/BCequal] (* clearly nonzero *)
    Factor[D[Lfn[TA, TB], {TA, 2}]] (* clearly positive *)
    Plot[Lfn[TA, (1 + Sqrt[1 + TA^2])/TA], {TA, 0.1, 8}]
    (* convex function, minimum at TA=Sqrt[3] *)

    tmp = Together[Simplify[D[Lfn[TA,(1 + Sqrt[1 + TA^2])/TA], TA]]]
    u = Sqrt[1 + TA^2];
    Simplify[tmp - (-2 + u)*(1 + u)^2/(TA^2*u)] (* 0 *)
    Simplify[tmp /. TA -> Sqrt[3]]. (* 0 *)

    ppu =ParametricPlot[{Lfn[TA,(1+Sqrt[1+TA^2])/TA]/dOfn[TA,(1+Sqrt[1+TA^2])/TA],
     1/dOfn[TA,(1+Sqrt[1+TA^2])/TA]},{TA,0.01,Sqrt[3]},PlotStyle->Red];
    ppl =ParametricPlot[{Lfn[TA,(1+Sqrt[1+TA^2])/TA]/dOfn[TA,(1+Sqrt[1+TA^2])/TA],
     1/dOfn[TA,(1+Sqrt[1+TA^2])/TA]},{TA,Sqrt[3],128},PlotStyle->Green];

    pairs[AB_] := {Lfn[AB[[1]], AB[[2]]]/dOfn[AB[[1]], AB[[2]]],
       1/dOfn[AB[[1]], AB[[2]]]};
    rdm[Npts_] :=
      Module[{k} ,
       Map[pairs,
        Table[{1 + RandomReal[{0, 32}], 1 + RandomReal[{0, 32}]}, {k, 1,
          Npts}]]];
    lp = ListPlot[rdm[20000], PlotRange -> All];
    rLdOtri = Show[{ppu, ppl, lp}, PlotRange -> All]

Figure 5: Triangles: $x=L/(2d_{O})$, $y=\rho/d_{O}$ with $\rho=1$.

The dots in the figure are from random choices for two of the $T$ values
(corresponding to two angles). The large number corresponding to large values
of $d_{O}$ is from using a uniform random distribution for the $T$ allowing
large values (so thin, or squat, triangles). The red curve is from tall thin
isosceles triangles. The green curve is from tall thin isosceles triangles.

###  16.1 Generalizing, angle-averaging

We have also used Mathematica Minimize on these tasks and it worked well for
$n=3$ and $n=4$ and NMinimize for larger $n$. However, having already proved
the result using Jensen’s inequality in §15.2 one learnt (very) little from
the exercise.

However, one way to learn a little more is to use the convexity of $\cot$ to
‘average two angles’ as remarked upon in the ‘tilting transformation’ treated
in §12. We keep $\rho$ fixed, say $\rho=1$. Choose two vertices, with angles
$\alpha_{i}$ and $\alpha_{j}$. Let

$\sigma_{i}=\tan(\frac{\alpha_{i}}{4})\qquad{\rm so\ }\ \
T_{i}=\frac{1-\sigma_{i}^{2}}{2\sigma_{i}}.$

Then, as

$\tan(\frac{\alpha_{i}+\alpha_{j}}{4})=\frac{\sigma_{i}+\sigma_{j}}{1-\sigma_{i}\sigma_{j}}).$

The convexity of the $\cot$ function on $(0,\pi/2)$ is

$\displaystyle 2\cot(\frac{\alpha_{i}+\alpha_{j}}{4})$ $\displaystyle\leq$
$\displaystyle\cot(\frac{\alpha_{i}}{2})+\cot(\frac{\alpha_{j}}{2}),$
$\displaystyle\frac{2(1-\sigma_{i}\sigma_{j})}{\sigma_{i}+\sigma_{j}}$
$\displaystyle\leq$ $\displaystyle T_{i}+T_{j}.$

From this, and the representation of $L$ as a sum of $T_{k}$, we see that $L$
is reduced by averaging two of the angles.

The same argument works for $i_{2}$ and for $i_{4}$ on using the convexity of
$\cot^{3}$ and of $\cot^{5}$.

It is also easy to work out the differences, by how much the quantities $L$,
$i_{2}$, etc. decrease.

    (* Write T[1]=1/tan(alpha1/2) and T[2] in terms of u1 = tan(alpha1/4) and u2 resp. *)
    T[1] = (1-u1^2)/(2*u1);
    T[2] = (1-u2^2)/(2*u2);
    (* After angle averaging/ optimal tilting when adjacent *)
    tAve= (1-u1*u2)/(u1+u2);
    Tout[1]= tAve;
    Tout[2]= tAve;

    Ldifference = Factor[2*(T[1]+T[2]-2*tAve)]
    (* (((u1 - u2)^2*(1 - u1*u2))/(u1*u2*(u1 + u2))) *)

    i2STfn[TT_]:= (TT+TT^3/3);
    i2difference = Factor[2*(i2STfn[T[1]]+i2STfn[T[2]]-2*i2STfn[tAve])]
    (* long expression - an obviously positive expression * Ldifference^3 *)

##  17 Tangential quadrilaterals

We begin with the context:

Figure 6:  wikipedia, attribution below

Attribution for Figure 6: By Alexgabi, jlipskoch
`https://commons.wikimedia.org/wiki/File:Laukien_sailkapena.svg,`
CC BY-SA 3.0
`https://commons.wikimedia.org/w/index.php?curid=34027107`

There is a huge literature on tangential quadrilaterals. See [18, 29, 30, 34,
36, 37, 38, 39, 62, 63]

In a tangential quadrilateral the two diagonals and the two tangency chords
are concurrent.

Theorem.Let $ABCD$ be a tangential quadrilateral and $O_{d}$ be the point of
intersection of its diagonals. Invert, using $O_{d}$ as pole, each of the
vertices, the inverse of $A$ denoted by $A_{1}$, etc. Let
$A_{1}B_{1}C_{1}D_{1}$ be the quadrilateral obtained by these inversions.
Then$A_{1}B_{1}C_{1}D_{1}$ is a tangential quadrilateral.

See [63]. Here is some additional comment. The intersection of the diagonals
of the two quadrilaterals coincide. Mobius maps map, in general, lines to
circles, but lines through the pole are preserved. Thus the angles between the
diagonals at $O_{d}$ are the same (as the diagonals are). The conformal map
$z\rightarrow\frac{1}{z}$ locally preserves angles between lines, and so does
its conjugate. I noticed some qualitative similarity between inputs and
outputs. In particular inputs like kites produced outputs like kites. We
already have that if the diagonals of the input cross at right angles so will
those of the output.

Figure 7: Inverse map. See [63].

The image under the inverse map with the incentre as pole seems often to take
tangential quadrilateral vertices to points that look somewhat like vertices
of a rhombus.

##  18 Bicentric polygons

In our calculations we often use $T_{k}$ as inout variables, or equivalently
$\eta_{k}=\rho\,T_{k}$. There are additional identities for bicentrics beyond
those that one has for tangential $n$-gons.

wikipedia gives, for bicentric quadrilaterals,

$None$

[71] studies bicentric hexagons and gives

$None$

There has been use of these in subsequent calculations.

The challenge put to me, admittedly in connection with $Q_{0-}$ rather than
the geometric functionals treated in this Part IIb, is: investigate Blaschke-
Santalo diagrams for tangential polygons. One triple that can be so
investigated is $(\rho,L,R_{V})$ for bicentric polygons. For triangles, see
the later section on Blundon’s inequality, e.g inequalities (19.1).

There is a huge literature on bicentric polygons dating back to the 19th
century. The topics include Fuss’s Theorem, Poncelet’s porism and more. Some
of this is impressive so is presented below - with nothing original of mine -
but with the hope that some might be of use in future attempts to produce
Blaschke-Santalo diagrams.

###  18.1 Fuss’s Theorem, Poncelet’s Porism

We would like to establish (polynomial) relations involving $(\rho,L,R_{V})$.
Fuss’s theorem(s) involves $(\rho,R_{V},d)$ where $d$ is the distance between
the circumcentre and incentre of the bicentric polygon. For triangles, The
quantity $d$ can be recognized in Blundon’s inequality (19.1) and is, as in
the first displayed equation below, $d=\sqrt{R_{V}(R_{V}-2\rho)}$.

The following is a (slightly adapted) quote from
`https://mathworld.wolfram.com/PonceletsPorism.html`

> The three numbers $(\rho,R_{V},d)$ will not be arbitrary and along with $n$,
> they will have to satisfy certain relations. For the case of a triangle, one
> such relation is sometimes called the Euler triangle formula:
>
> $R_{V}^{2}-2R_{V}\rho-d^{2}=0.$
>
> One of popular notations for such relations (which is necessary and
> sufficient for existence of a bicentric polygon) can be given in terms of
> the quantities
>
> $a=\frac{1}{R_{V}+d},\ \ b=\frac{1}{R_{V}-d},\ \ c=\frac{1}{\rho}.$
>
> For a triangle, the Euler formula has the form:
>
> $a+b=c,$
>
> for a bicentric quadrilateral
>
> $a^{2}+b^{2}=c^{2}.$
>
> The relationship for a bicentric pentagon is
>
> $4(a^{3}+b^{3}+c^{3})=(a+b+c)^{3}.$
>
> Let
>
> $\displaystyle E_{1}$ $\displaystyle=$ $\displaystyle-a^{2}+b^{2}+c^{2},$
> $\displaystyle E_{2}$ $\displaystyle=$ $\displaystyle a^{2}-b^{2}+c^{2},$
> $\displaystyle E_{3}$ $\displaystyle=$ $\displaystyle a^{2}+b^{2}-c^{2}.$
>
> The relationship for a bicentric hexagon is
>
> $\frac{1}{E_{1}}+\frac{1}{E_{2}}=\frac{1}{E_{3}}$

###  18.2 Triangles, again

The sides $s_{1}$, $s_{2}$, $s_{3}$ of a triangle are the roots of the cubic

$y^{3}-Ly^{2}+(\frac{L^{2}}{4}+\rho^{2}+4\rho\,R_{V})y-2L\rho R_{V}=0.$ (18.1)

ToDo. Find conditions on the coefficients of this cubic in order that it have
3 positive roots with the largest root less than the sum of the other two.
Sturm sequences might be useful. Do Blundon’s inequalities come from this?

###  18.3 Bicentric quadrilaterals

The wikipedia article ‘Bicentric quadrilateral’ contains many items. There is
a quartic equation with coefficients in terms of $L$, $\rho$ and $R_{V}$ whose
solutions are the sides of a bicentric polygon:

$y^{4}-Ly^{3}+(\frac{L^{2}}{4}+2\rho^{2}+2\rho{\sqrt{4R_{V}^{2}+\rho^{2}}})y^{2}-\rho
L({\sqrt{4R_{V}^{2}+\rho^{2}}}+\rho)y+\rho^{2}\frac{L^{2}}{4}=0.$ (18.2)

ToDo. Find conditions on the coefficients of this quartic in order that it
have 3 positive roots with the largest root less than the sum of the other
three. Sturm sequences might be useful.

There are huge number of inequalities.

Theorem.If a bicentric quadrilateral has an incircle and a vertex-circumcircle
with radii $\rho$ and $R_{v}$ respectively, then its area $A$ satisfies

$\frac{1}{2}\rho\,L=A\geq 2\rho\sqrt{2\rho(\sqrt{4R_{v}^{2}+\rho^{2}}-\rho)},$

where equality holds if, and only if, the quadrilateral is also an isosceles
trapezium.
See [36].
For a square $\rho=1$, $R=R_{v}=\sqrt{2}$, $A=4$ gives equality in the
preceding inquality.

For a bicentric quadrilateral

$\frac{1}{32}(L^{2}-16A)\leq R_{v}^{2}-\rho^{2}$

with equality if and only if the hexagon is regular.

###  18.4 Bicentric hexagons

For a bicentric hexagon

$\frac{1}{36}(L^{2}-8\sqrt{3}\,A)\leq R_{V}^{2}-\frac{4}{3}\rho^{2}$

with equality if and only if the hexagon is regular.

For bicentric $n$-gons see [59].

###  18.5 $R=R_{v}=d_{O}$ for regular polygons

In a regular $n$-gon, the side $s_{n}=2\rho\tau_{n}$ where
$\tau_{n}=\tan(\pi/n)$. The circumradius $R$ and $d_{O}$ coincide. We have

$4(R^{2}-\rho^{2})=s_{n}^{2}$

so

$R=d_{O}=\sqrt{\rho^{2}+\frac{s_{n}^{2}}{4}}=\rho\sqrt{1+\tau_{n}^{2}}.$

See Part IIa §8 for the formulae for $L=i_{0}$, $i_{2}$ and $i_{4}$.

###  18.6 Bicentrics from regular

Given any regular $n$-gon any choice of 3 of its vertices gives a bicentric
polygon as all triangles are bicentric.

Given a regular 8-gon, selection of its alternate vertices gives a bicentric
4-gon, a square.

Given a regular 9-gon, there is a selection of 5 of its vertices giving a
bicentric 5-gon. See [86, 87].

##  19 Blaschke-Santalo diagrams for tangential polygons

###  19.1 Definitions for Blaschke-Santalo diagrams

Where we feel it helps the exposition there will be some repetition of
material already presented in PartIIa. We always choose the origin of our
coordinate system to be at the incentre. The inradius $\rho$ and perimeter $L$
will occur in our diagrams, and there will be various choices of a third
geometric quantity which, for the present, we denote by $q$. In our Blaschke-
Santalo diagrams the horizontal axis is usually $x=L/q$ and the vertical axis
$y=\rho/q$. (Further work in which $x=L/\rho$ and, as before, $y=\rho/q$ might
be undertaken, motivated by having the same $x$ for the different $q$, e.g.
$q$ being $i_{2}^{1/3}$, $Q_{0-}^{1/4}$, etc.) The area of a tangential
polygon is $A=\rho\,L/2$. The quantity $x/y=L/\rho$ is denoted by $B$ in [69]
. We have $B\geq{2\pi}$ with equality only for the disk, and for any triangle,
$B\geq{6\sqrt{3}}$): see [1]. The perimeter and inradius of a regular $n$-gon
are related by

$\rho=\frac{L}{2n}\cot(\frac{\pi}{n}),$

and for a tangential $n$-gon

$B\geq{2n\tan(\pi/n)}.$

The other domain functionals are as follows.
(1) There is a first set of geometric quantities:
$\bullet$ circumradius $R$ (radius of the smallest disk containing the
region);
$\bullet$ the distance from the incentre to the boundary $d_{0}$;
(2) There are moments about the incentre:
$\bullet$ the second boundary moment $i_{2}$;
$\bullet$ the fourth boundary moment $i_{4}$;
(3) There are quantities leading to $Q_{0,-}$:
$\bullet$ $\Sigma_{\infty}=\rho\,i_{2}/16$;
$\bullet$ $0>\Sigma_{1}=(i_{2}^{2}/L-i_{4})/16$;
$\bullet$ $Q_{0-}$ the lower bound on torsional rigidity treated in Part I,
i.e. [48].
We defer discussion of these last items, $Q_{0,-}$, until Part III.

In connection with bicentric polygons, including triangles, the radius of the
circle through the vertices is denoted $R_{V}$, and we always have
$R\leq{R_{V}}$.

An outline of the remainder of this section follows.

  * •

In §19.2 we define cap domains which enter the account of the $(\rho,L,R)$
diagram in the next subsection.

  * •

In §19.3 we present formulae for $d_{O}$ for various tangential $n$-gons.

  * •

In §19.4 we begin by noting that the account for general convex domains in
[10] has on one of its boundaries the 2-cap. We have done some computations,
and have a belief that regular $n$-gons occur on one of the boundaries.
However, the circumradius $R$ doesn’t seem to be a good lead-in for
computations related to ($i_{2}$, $i_{4}$ and) $Q_{0-}$.

  * •

§19.5 is the main subsection. The quantity $d_{O}$ is a function easily
defined in terms of the $T_{k}$ occuring in expressions for $L$, $i_{2}$ and
$i_{4}$, and hence in $Q_{0-}$. The expressions for $L$, $i_{2}$ and $i_{4}$
are given in Part IIa. The only Blaschke-Santalo diagrams presented here are
those for $(\rho,L,d_{O})$. Future work may involve $(\rho,L,i_{2})$ and
$(\rho,L,i_{2})$, but for now there is just occasional comment on relevant
inequalities for particular tangential $n$-gons.

  * •

Though we believe it to be an aside to our main purpose of investigating
functionals related to $Q_{0-}$ over all tangential polygons, in §19.6 we
propose to treat, for bicentric polygons the triple $(\rho,L,R_{V})$. So far,
the main result is just that already in the literature for triangles. Some
items in §18 may be useful in future efforts.

###  19.2 Some extremals amongst circumgons: the 1-cap and symmetric 2-cap

Because our long-term motivation concerns $Q_{0-}$ and this involves $i_{2}$
(and $i_{4}$) se begin with the following.

Let $\rho>0$. Among all plane convex sets $\Omega$ with fixed positive area
$A$ (with $A>\pi\rho^{2}$) that contain a disk of radius $\rho$ around the
origin, $i_{2}$ is maximized if $\Omega$ is the single-cap, the convex hull or
a disk of radius $\rho$ and a point (the point being, up to rotation
invariance, uniquely defined by the area constraint).
The proof is elementary. See [35] where it is a Lemma needed prior to
establishing gradient bounds for the torsion function. Exactly the same proof
gives the corresponding result for $i_{4}$.

Consider tangential polygons with $\rho$ and $d_{O}>\rho$ fixed. Amongst these
the 1-cap
$\bullet$ minimizes $A$, $L$, $i_{2}$ and $i_{4}$ and
$\bullet$ minimizes $Q_{0}$.
This is because of domain monotonicity of these functionals.
Question. Given two tangential polygons $\Omega_{0}$ and $\Omega_{1}$ (with
the same incentre? and) with $\Omega_{0}\subset\Omega_{1}$, is
$Q_{0-}(\Omega_{0})\leq Q_{0-}(\Omega_{1})$?

Figure 8: Left: the symmetric 2–cap. Right: the 1-cap.

We can easily calculate the perimeter of the 1-cap. Denote by $\alpha$ the
angle at its vertex on the circle radius $d_{O}$. Then
$\sin(\alpha/2)=\rho/d_{O}$ and

$\displaystyle L$ $\displaystyle=$
$\displaystyle(\pi+\alpha)\rho+2\sqrt{d_{O}^{2}-\rho^{2}},$
$\displaystyle\frac{L}{d_{O}}$ $\displaystyle=$
$\displaystyle\left(\pi+2\arcsin(\frac{\rho}{d_{O}})\right)\,\frac{\rho}{d_{O}}+\sqrt{1-(\frac{\rho}{d_{O}})^{2}}.$

We will see this as occuring as the lower right boundary in diagrams with
$x=L/d_{O}$, $y=\rho/d_{O}$. The limiting cases are

  * •

$d_{O}\rightarrow\rho$, $y\rightarrow{1}$, $x\rightarrow{2\pi}$ corresponding
to a disk;

  * •

$d_{O}\rightarrow\infty$, $y\rightarrow{0}$, $x\rightarrow{2}$ corresponding
to a long flat shapes like the 1-cap.

  * •

$d_{O}\rightarrow\infty$, $y\rightarrow{0}$, $x\rightarrow{4}$ corresponding
to a long flat shapes like the 2-cap.

  * •

The quadrilateral might have one side tending to zero, say as a symmetric
tangential trapezium tending to an isosceles triangle. When the squat
isosceles triangle becomes very thin, $y\rightarrow{0}$, $x\rightarrow{4}$.

We will see the 2-cap in §19.4.

###  19.3 $d_{O}$ for various tangential $n$-gons

The distances from the incentre of the vertices of a tangential $n$-gon are
given by

$\sqrt{\rho^{2}+\eta_{k}^{2}}=\rho\sqrt{1+T_{k}^{2}}.$

####  19.3.1 $R$, $d_{O}$ for regular polygons

In a regular $n$-gon, the side $s_{n}=2\rho\tau_{n}$ where
$\tau_{n}=\tan(\pi/n)$. The circumradius $R$ and $d_{O}$ coincide. We have

$4(R^{2}-\rho^{2})=s_{n}^{2}$

so

$R=d_{O}=\sqrt{\rho^{2}+\frac{s_{n}^{2}}{4}}=\rho\sqrt{1+\tau_{n}^{2}}.$

See Part IIa §8 for the formulae for $L=i_{0}$, $i_{2}$ and $i_{4}$.

####  19.3.2 $R$, $d_{O}$ for triangles

For an equilateral triangle

$L=6\sqrt{3}\rho,\qquad R=d_{O}=2\rho.$

In general $R$ and $d_{O}$ differ. The circumradius of an acute angled
triangle is the radius of the circle through the three vertices ($R=R_{V}$).
For a triangle with one interior angle measuring more than $\pi/2$, an obtuse
triangle, the circumradius is half the length of the longest side (and
$R<R_{V}$).

For triangles equation (15.4) is

$1-\frac{1}{T_{1}T_{2}}-\frac{1}{T_{2}T_{3}}-\frac{1}{T_{3}T_{1}}=0.$

As in Part IIa, most of our calculations have, to date, concentrated on
isosceles triangles. See Part IIa §9 for the formulae for $L=i_{0}$, $i_{2}$
and $i_{4}$. We return to triangles in §19.5.1. For now it is sufficient to
record that

$None$

####  19.3.3 $d_{O}$ for tangential quadrilaterals

For tangential quadrilaterals equation (15.5) is

$\frac{1}{T_{1}}+\frac{1}{T_{2}}+\frac{1}{T_{3}}+\frac{1}{T_{4}}-\frac{1}{T_{1}T_{2}T_{3}}-\frac{1}{T_{1}T_{2}T_{4}}-\frac{1}{T_{1}T_{3}T_{4}}-\frac{1}{T_{2}T_{3}T_{4}}=0.$

We return to tangential quadrilaterals in §19.5.2. For now it is sufficient to
record that

$None$

###  19.4 $(\rho,L,R)$

For any plane convex set

$\rho\leq{R},\ L\leq{2\pi\,R},\ 2\pi\rho\leq{L},\ 4R\leq{L}.$

There is some discussion of how to compute $R$ at
`https://mathematica.stackexchange.com/questions/121987/how-to-find-the-
incircle-and-circumcircle-for-an-irregular-polygon`

We can scale our shapes so there is no loss of generality in setting $\rho=1$.
Define

$y=\frac{1}{R}\ \ {\rm and}\ \ x=\frac{L}{R},$

(with our $x$ a factor of $2\pi$ greater than than in [10]).

$x=2\pi,\ y=1\qquad{\rm for\ a\ disk}.$

[10] establish, for convex sets,

$4\left(\sqrt{1-y^{2}}+\ y\,{{\rm arcsin}(y)}\right)\leq x=\frac{L}{R}\leq
4\left(\sqrt{1-y^{2}}+{\rm arcsin}(y)\right).$

The lower bound at the left corresponds to values for a symmetric 2-cap, which
is a circumgon tangential polygon. However, the right-hand, upper bound is for
a convex shape which is not a tangential polygon. Our computations suggest
that amongst the tangential polygon shapes which will occur on the right-hand
upper bound are the regular polygons.

In Figure 9 the right-most blue curve is the shape from [10] which is not a
tangential polygon. The upper-left green curve is form the symmetric 2-cap.
The red dots are some regular polygons, the rightmost upper dot the circular
disk. The lowest is the equilateral triangle, and the next up, joining the
upper blue curve is the square. The scatter of blue dots are from tangential
quadrilaterals, and the upper blue line from rhombi.

Figure 9: Except for the lower curve, some results on $(L/R,\rho/R)$ pairs for
tangential polygons. See description in text.

###  19.5 $(\rho,L,d_{O})$

####  19.5.1 Triangles

See also PartIIa §9.

Amongst all isosceles triangles with given inradius

  * •

the equilateral triangle minimizes $d_{O}$,

  * •

the equilateral triangle minimizes $A$ and $L=2A/\rho$ and $L^{2}/A$, and

  * •

the equilateral triangle minimizes $i_{2}$ and $i_{2}/(A\,L)$.

At given $\rho$ and $A$ greater than the area of the equilateral triangle of
the same inradius, amongst isosceles triangles

  * •

$d_{O}$ is maximised by tall isosceles triangles (apex angle $\alpha<\pi/3$
small, $\sigma<2-\sqrt{3}$ small)

  * •

$d_{O}$ is minimised by short squat isosceles triangles (apex angle
$\alpha>\pi/3$ near $\pi$, $\sigma>2-\sqrt{3}$ near 1).

(Above may be true for all triangles.)

The calculations for $i_{2}$ for the following are yet to be done.
Conjecture At given $\rho$ and $A$ greater than the area of the equilateral
triangle of the same inradius, amongst isosceles triangles

  * •

$i_{2}$ is maximised by tall isosceles triangles (apex angle $\alpha<\pi/3$
small, $\sigma<2-\sqrt{3}$ small)

  * •

$d_{O}$ is minimised by short squat isosceles triangles (apex angle
$\alpha>\pi/3$ near $\pi$, $\sigma>2-\sqrt{3}$ near 1).

####  19.5.2 Tangential quadrilaterals

Rhombi

Amongst all rhombi with given side,

  * •

the square minimizes $d_{O}$,

  * •

the square maximizes $A$ and $\rho=2A/L$, and

  * •

the square minimizes $i_{2}/(A\,L)$.

These are easily established, as follows.

$A=2\rho^{2}(T+\frac{1}{T})\qquad L=4\rho(T+\frac{1}{T}).$

Hence

$\frac{L^{2}}{A}=8(T+\frac{1}{T}),$

which is minimized when $T=1$, the square.
Next

$\displaystyle i_{2}$ $\displaystyle=$ $\displaystyle
2\rho^{3}\left(T+\frac{1}{T}+\frac{1}{3}(T^{3}+\frac{1}{T^{3}})\right),$
$\displaystyle=$ $\displaystyle\frac{2\rho^{3}}{3}(T+\frac{1}{T})^{3},$

so that

$\frac{i_{2}}{AL}=\frac{1}{12}(T+\frac{1}{T}),$

which is minimized at $T=1$.

We also have the following.

Amongst all rhombi with given inradius

  * •

the square minimizes $d_{O}$,

  * •

the square minimizes $A$ and $L=2A/\rho$ and $L^{2}/A$, and

  * •

the square minimizes $i_{2}$ and $i_{2}/(A\,L)$.

A plot of $y=\rho/d_{O}$ against $x=L/d_{O}$ is shown in Figure 10

Figure 10: For rhombi, plot of $y=\rho/d_{O}$ against $x=L/d_{O}$. The top
right corner is $(1/\sqrt{2},4\sqrt{2})$ corresponding to a square.

Kites

Amongst kites with given area $A$ and distance $d$ between apexes, the rhombus
has

  * •

the smallest perimeter $L$,

  * •

the largest inradius $\rho=2A/L$, and

  * •

the smallest $4I_{2}=\rho\,i_{2}$.

A proof uses Steiner symmetrisation.

One can easily recover results which, in a more general form, are given in the
next subsubsection.

Amongst kites with given sides $s_{1}$ and $s_{2}$ the right kite has the
largest area. For ease of writing, suppose $s_{1}<s_{2}$. The rhombus case can
be treated separately. Let $h$ be the distance between the vertices adjacent
to unequal sides. Let $\alpha_{1}$ be the angle at the vertex on both sides of
length $s_{1}$. Denote by $\beta$, the repeated angle, the angle at vertices
adjacent to the unequal sides.

$A=\frac{1}{2}dh\ {\rm which\ can\ be\ written\ }\ A=s_{1}s_{2}\sin(\beta).$

The formula at the right can be deduced from that on the left using the
trigonometry:

$h=2s_{1}\sin(\alpha_{1}/2),\qquad
d=s_{2}\frac{\sin(\beta)}{\sin(\alpha_{1}/2)},$

the latter from the triangle sine rule. Thus since $A=s_{1}s_{2}\sin(\beta)$,
the area is maximized at $\beta=\pi/2$.

We remark

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle\frac{s_{1}s_{2}\sin(\beta)}{s_{1}+s_{2}},$
$\displaystyle\frac{L}{\rho}$ $\displaystyle=$
$\displaystyle\frac{2(s_{1}+s_{2})^{2}}{s_{1}s_{2}\sin(\beta)}.$

Also, on using $s_{1}<s_{2}$,

$\frac{d_{O}}{\rho}=\sqrt{1+{\rm
max}(\cot(\frac{\beta}{2}),\cot(\frac{\alpha_{1}}{2}))^{2}},$

and the sine rule for triangles gives $\alpha_{1}$ in terms of $s_{1}$,
$s_{2}$ and $\beta$.

Other tangential quadrilaterals

For a quadrilateral with angles $\alpha_{k}$ and sequence of sides
$[s_{1},s_{2},L/2-s_{1},L/2-s_{2}]$, the area is

$A=\sqrt{s_{1}\,s_{2}\,(L/2-s_{1})\,(L/2-s_{2})}\sin(\frac{\alpha_{1}+\alpha_{3}}{2}).$

For any quadrilateral, since the sum of the angles is $2\pi$,

$\sin(\frac{\alpha_{1}+\alpha_{3}}{2})=\sin(\frac{\alpha_{2}+\alpha_{4}}{2})).$

The formula for the area $A$ above gives the following.

Amongst all tangential quadrilaterals with given sequence of sides, the
bicentric quadrilateral maximizes $A$ and $\rho=2A/L$.
A kite is bicentric iff it is a right kite.

In general, amongst all kites with a given pair of sides it is not the case
that $i_{2}/(A\,L)$ is minimized by a right kite. (Let $\beta$ be the repeated
angle in the kite. The quantity $i_{2}/(A\,L)$ plotted as a function of
$1/\tan(\beta/2)$ appears to have a unique minimum which for a rhombus is at
$\beta=\pi/2$, but in general is not.)

More on quadrilaterals

[53] treat quadrilaterals with diagonals intersecting at right angles. If the
side lengths are denoted $s_{1}$, $s_{2}$, $s_{3}$, $s_{4}$ these satisfy

$s_{1}^{2}+s_{3}^{2}=s_{2}^{2}+s_{4}^{2}.$

If a quadrilateral is tangential and has its diagonals intersecting at right
angles it is kite. (See also [52]

Bicentric quadrilaterals, $(\rho,L,d_{O})$

Figure 11: $x=L/d_{O}$, $y=\rho/d_{O}$ for bicentric quadrilaterals

The upper left boundary curve in Figure 11 (red) corresponds to right kites:
the lower right curve (green) corresponds to tangential isosceles trapeziums.
The dots are just from randomly generated tangential quadrilaterals. The
uppermost point $(4\sqrt{2},1/\sqrt{2})$ corresponds to a square. The bottom
boundary is approached by long thin shapes, and the left-most point is $(2,0)$
corresponding to the limit of thin shapes where $L\sim{2d_{O}}$.

Another approach to the boundary curves involves working with the tangent
lengths $\eta_{j}$. For a bicentric quadrilateral with $\rho=1$ we have

$\eta_{1}\eta_{3}=\rho^{2}=1=\eta_{2}\eta_{4}\ {\rm so}\ \
L=\eta_{1}+\frac{1}{\eta_{1}}+\eta_{2}+\frac{1}{\eta_{2}}.$

Choose $\eta_{1}$ to be the largest of the $\eta_{k}$, so $\eta_{1}>1$ and
$d_{O}=\sqrt{1+\eta_{1}^{2}}$. Now $\eta_{2}$ must be less than or equal to
$\eta_{1}$ and greater than or equal to $1/\eta_{1}$, and consequently So

$L_{-}=2+\eta_{1}+\frac{1}{\eta_{1}}\leq L\leq
2(\eta_{1}+\frac{1}{\eta_{1}})=L_{+}.$

Plotting $y=1/d_{O}$ against $x=L_{-}/d_{O}$ gives one of the boundary curves,
and the other is from $y=1/d_{O}$ against $x=L_{+}/d_{O}$.

As an aside we remark that for a bicentric quadrilateral for which $\rho=1$
and $s_{1}$ ia the largest side (so $s_{1}\geq{2}$) and $s_{2}$ is the larger
of the other pair (so $s_{2}>1$),

$L=\frac{s_{1}s_{2}(s_{1}+s_{2}+\sqrt{4+(s1-s2)^{2}}}{s_{1}s_{2}-1}.$

General tangential quadrilaterals

In Figure 12 the scatter of (blue) points towards the left upper are from
$T_{k}$ values with the first 3 not all that far from 1. (Those blue dots that
are near the bottom come from those with $T_{4}$ large, such as occurs if the
first 3 of the $T_{k}$ were near $1/\sqrt{3}$, the $\alpha_{k}$ near
$2\pi/3$.) The red dots are from choosing the first 3 $T_{k}$ randomly in the
interval $(0,1000)$. Large values of $T_{k}$ cause $d_{O}$ to be large, hence
the cluster close to $y=0$. We haven’t definitively defined the boundaries.
However we suspect the upper left boundary, at least for small enough $y$
($d_{O}$ large) have the minimum $L$ shapes somewhat like the 1-cap, perhaps
kites with a small apex angle at $d_{O}$.

Figure 12: $x=L/d_{O}$, $y=\rho/d_{O}$ for tangential quadrilaterals

The dashed curve is the bound from the 1-cap, discussed nearthe beginning of
this section.

####  19.5.3 Tangential pentagons

Bicentric pentagons

An elaborate formula for the area of a cyclic pentagon in terms of side
lengths is given in [76].

####  19.5.4 Tangential hexagons

Bicentric hexagons

A question that arises is, what conditions other than

$s_{1}+s_{3}+s_{5}=s_{2}+s_{4}+s_{6}=\frac{L}{2},$

are satisfied by the sides, or tangent lengths $\eta_{j}$, for a bicentric
hexagon. We have, for tangential hexagons,

$s_{1}=\eta_{1}+\eta_{2},\ s_{2}=\eta_{2}+\eta_{3},\ s_{3}=\eta_{3}+\eta_{4},$
$s_{4}=\eta_{4}+\eta_{5},\ s_{5}=\eta_{5}+\eta_{6},\ s_{6}=\eta_{6}+\eta_{1},$

or, in the notation of §15.1,

$M(6)\,{\mathbf{\eta}}={\mathbf{s}}.$

From [73] equations (2.29), (2.30)

$\rho\sqrt{\frac{\eta_{1}\eta_{3}\eta_{5}}{\eta_{1}+\eta_{3}+\eta_{5}}}=\eta_{2}\eta_{5}=\eta_{1}\eta_{4}=\eta_{3}\eta_{6},$

and, from [73] equation (2.26)

$\eta_{1}\eta_{3}+\eta_{3}\eta_{5}+\eta_{5}\eta_{1}=\eta_{2}\eta_{4}+\eta_{4}\eta_{6}+\eta_{6}\eta_{2}=\rho^{2}.$

Bicentric hexagons An elaborate formula for the area of a cyclic hexagon in
terms of side lengths is given in [76].

###  19.6 $(\rho,R_{V},L)$ and generalizing Blundon’s inequality

We hope to treat bicentric quadrilaterals in the future, but, for now, here
are results for triangles.

####  19.6.1 Triangles

One of the many entries in wikipedia’s list of triangle inequalities is the
following, which in much of the literature is known as Blundon’s inequality:

$2R_{V}^{2}+10R_{V}\rho-\rho^{2}-2\sqrt{R_{V}}(R_{V}-2\rho)^{3/2}\leq\frac{L^{2}}{4}\leq
2R_{V}^{2}+10R_{V}\rho-\rho^{2}+2\sqrt{R_{V}}(R_{V}-2\rho)^{3/2}.$ (19.1)

Quoting from wikipedia:

> Here the expression
>
> $\sqrt{R_{V}^{2}-2R_{V}\rho}={\rm dist(incentre,circumcentre)},$
>
> In the double inequality (19.1), the first part holds with equality if and
> only if the triangle is isosceles with an apex angle of at least $\pi/3$,
> and the last part holds with equality if and only if the triangle is
> isosceles with an apex angle of at most $\pi/3$. Thus both are equalities if
> and only if the triangle is equilateral.

Let $x=L/R_{V}$, $y=\rho/R_{V}$. The allowed values of the $(x,y)$ pair is the
region inside the curves given by

$4(2+10y-y^{2}-2(1-2y)^{3/2})\leq x^{2}\leq 4(2+10y-y^{2}+2(1-2y)^{3/2}),$

and shown in Figure 13.

Figure 13: The lhs of the Blunden inequality is shown in red, the rhs in blue
dashed.

Some of the behaviour of isosceles triangles for the various Blaschke-Santalo
diagrams is indicated in the table below.

| squat | tall
---|---|---
$L/R$ | $\sim{4}$ | $\sim{4}$
$\rho/R$ | $\rightarrow{0}$ | $\rightarrow{0}$
$L/R$ | $\sim{4}$ | $\sim{2}$
$\rho/R$ | $\rightarrow{0}$ | $\rightarrow{0}$
$L/R_{V}$ | $\sim{0}$ | $\sim{4}$
$\rho/R_{V}$ | $\rightarrow{0}$ | $\rightarrow{0}$

##  20 Moments about the centroid, $I_{c}$

The moment of inertia about the centroid, $I_{c}$ is, amongst $n$-gons with a
given area, minimized by the regular $n$-gon:

$\displaystyle I_{c}$ $\displaystyle\geq$ $\displaystyle I_{c}({\rm reg-n-
gon}),$ $\displaystyle=$ $\displaystyle\frac{1}{4}\rho({\rm reg-n-
gon})\,i_{2}({\rm reg-n-gon}),$ $\displaystyle=$
$\displaystyle\frac{A^{2}(3+\tau_{n}^{2})}{6n\tau_{n}}\ {\rm where\ }\
\tau_{n}=\tan(\frac{\pi}{n}).$

(The calculations for the equations above are in Part IIa.)

The St Venant inequality is the left-hand side below

$\frac{Q_{0}}{A^{2}}\leq\frac{1}{8\pi}\leq\frac{I_{c}}{4A^{2}}\leq\frac{I_{2}}{4A^{2}},$

where, as before, $I_{2}$ is the moment about the incentre, and, for
tangential polygons, $I_{2}=\rho i_{2}/4$.

We can add this inequality into that near the end of Part I §3.2 that for any
tangential polygon

$\frac{1}{8}\rho^{2}A\leq Q_{0-}\leq
Q_{0}\leq\frac{1}{4}I_{c}\leq\frac{1}{4}I_{2}\leq Q_{0+}.$

###  20.1 Formulae for $I_{2}-I_{c}=A|z_{c}-z_{I}|^{2}$

The identity in the subsection heading is from the Parallel Axis Theorem.
`https://en.wikipedia.org/wiki/Parallel_axis_theorem` Denote by $z_{c}$ the
coordinates of the centroid, and $z_{I}$ those of the incentre. Take, as
elsewhere, the incentre to be the origin. Write

$d_{IG}=|z_{c}-z_{I}|.$

Triangles

For triangles $d_{IG}$ is given in terms of the side lengths in
`https://mathworld.wolfram.com/Incenter.html`
Let the sides be denoted by $s_{k}$ and define

$\displaystyle S_{1}$ $\displaystyle=$ $\displaystyle s_{1}+s_{2}+s_{3},$
$\displaystyle S_{2}$ $\displaystyle=$ $\displaystyle
s_{1}s_{2}+s_{2}s_{3}+s_{3}s_{1},$ $\displaystyle S_{3}$ $\displaystyle=$
$\displaystyle s_{1}s_{2}s_{3}.$

Then

$d_{IG}^{2}=\frac{5S_{1}S_{2}-S_{1}^{3}-18S_{3}}{9S_{1}}.$ (20.1)

Quadrilaterals

For a rhombus the centroid and incentre coincide at the point of intersecion
of the diagonals.

We have yet to derive, for tangential quadrilaterals or special cases of
these, formulae analogous to (20.1).

See [53, 52] for results concerning centroids, and characterizations of kites,
etc. See also [65, 31].

Part III: Other inequalities and properties for $Q_{0}$,
isoperimetric inequalities, etc.

## Abstract for Part III

A referee asked for a bit more on ‘context’ and more numerics. There is a huge
literature on bounds for torsional rigidity and in this part I focus on that
most relevant to convex polygons.

  * •

I also return to more on triangles, especially isosceles triangles for which,
unsuprisingly, if one makes use of information specific to triangles, one can
improve on my $Q_{0-}$ bound of Part I.

  * •

Numerics for rhombi again indicate that $Q_{0-}$ is quite close to the actual
torsional rigidity.

##  21 Introduction to Part III

A famous open problem is as follows.
Amongst all simple polygons with $n$ vertices with a given area, does the
regular $n$-gon have the greatest torsional rigidity?
[69] establishes this for $n=3$ and $n=4$ but the question is open for $n>4$.
It remains open for tangential $n$-gons too.
It will be easier to investigate, for tangential $n$-gons, this for the lower
bound $Q_{0,c}$ rather than the actual torsional rigidity $Q_{0}$. The results
in this direction are, so far, very slight (just isoceles triangles). We have
yet to establish it for general triangles $n=3$.

An outline for this part is as follows.

  * •

In §22 we present some previously published bounds. We also introduce the
style of isoperimetric inequalities which will be studied in later sections.

  * •

In §23 we review some bounds on $Q_{0}$.

  * •

In §24 we study isosceles triangles There are many questions on how the
geometry affects the domain functionals. We have, in §24.1, some isoperimetric
results: some functionals are optimized, over isosceles triangles with given
area, by the equilateral triangle.

  * •

In §25 (yet to be written!) we will, very briefly, consider extending the work
on $Q_{0-}$ to general triangles.

  * •

In §26 we show, numerically for rhombi, how close $Q_{0-}$ is to $Q_{0}$.
This parallels the work on isosceles triangles presented at the end of Part I.

  * •

In §27 we consider Blaschke-Santalo diagrams involving $Q_{0}$, or $Q_{0-}$,
for tangential polygons.

  * •

In §28 I give further questions in connection with $Q_{0-}$. Some of these are
of the form: if some property has been established for $Q_{0}$, does $Q_{0-}$
have the same property. Domain monotonicity is one such property, and one
question is whether it remains, for $Q_{0-}$ under corner cutting (see Part
IIb §12). Another, not discussed there, is whether, amongst tangential
$n$-gons with a given inradius, the regular $n$-gon minimizes $Q_{0-}$.

##  22 Bounds on $Q_{0}$, esp. isoperimetric

Other papers involving the torsional rigidity of tangential polygons include
[20]§2.2 involving web functions and [78]. Web functions are particularly
simple for tangential polygons, and align with the similar level curves of
[69], level curves which are the same shape as the boundary. In [78] the
bounds are in terms of

$I(q,\partial\Omega)=\int_{\Omega}{\rm dist}(z,\partial\Omega)^{q}.$

(The notation is that of [42] as, in this Supplement, there are already other
uses for $I$ and $i$. The capital $I$ is, as with our other use, an integral
over the domain.) For tangential polygons $I_{0}(\partial\Omega)=|\Omega|$ and

$I(q,\partial\Omega)=\frac{(p+1)(p+2)}{(q+1)(q+2)}I(p,\partial\Omega)\rho(\Omega)^{q-p}=\frac{1}{(q+1)(q+2)}\,|\Omega|\,\rho(\Omega)^{q}.$

We specialise a much more general theorem of [78] to the following.
Theorem [78]. Let $\Omega$ be a tangential polygon. Then

$Q_{0}(\Omega)\geq\frac{1}{16}(p+1)(p+2)I(p,\partial\Omega)\,\rho(\Omega)^{2-p}\qquad{\rm
where}\ -1\leq p<\infty.$

Equality holds if $\Omega$ is a disk.
In particular, with $p=2$

$\frac{1}{2}|\Omega|\rho^{2}=3I(2,\partial\Omega)\leq 4Q_{0}(\Omega),$

which recovers [69]§5.8 equation(7) on p100. See inequality (1.4) in Part I.
[78] is largely concerned with how this might generalize to domains other than
tangential polygons.

The famous open problem stated at the beginning of this part is as follows.
Amongst all simple polygons with $n$ vertices with a given area, does the
regular $n$-gon have the greatest torsional rigidity?
Repeating from there: [69] establishes this for $n=3$ and $n=4$ but the
question is open for $n>4$.