Of course the question can be asked with a smaller sets, e.g. convex polygons
or tangential polygons or bicentric polygons or … To date there are no answers
for the torsional rigidity. There are, however, for some other domain
functionals: see Part IIb for purely geometric ones. The conformal inradius
and related radii are others that on fixing the area is extreme for the
regular $n$-gon (see [83]). 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 6 below presents some (repeating some entries from Table 4 of Part IIb):

${\cal D}_{1}$ |  | ${\cal D}_{2}$ | Remark
---|---|---|---
area | min | perimeter |
area | max | inradius | $A=\rho L/2$
inradius | min | perimeter | Jensen inquality
inradius | min | area | ”
inradius | min | $Q_{0}$ | Solynin [82, 84]
|  |  | see below
Table 6: Tangential polygons

In [82] the result, at fixed inradius $\rho$, that $Q_{0}$ is minimized at the
regular $n$-gon is first given, in Theorem 1, for tangential $n$-gons, and,
after that, in Theorem 2, for more general $n$-gons. (I have yet to check the
proofs.)

###  22.1 Bounds on $Q_{0}$ for convex $\Omega$

We have, in Part I, reported the results, from [69]

$\frac{A^{2}}{4B}\leq Q_{0}\leq\frac{A^{2}}{8\pi}.$

The left-hand inequality is in Part I, inequality (1.4): the right-hand
inequality is the St Venant Inequality. There is equality in both for disks.
[69] p99 gives, for convex domains,

$B\leq\frac{2A}{\rho^{2}},$

which is an equality for tangential polygons. Combining these gives for convex
domains, the left-hand inequality in

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

The left hand inequality is from [69]§5.8 equation (7), and equality occurs
for $\Omega$ a disk. The right hand inequality with $c=\frac{1}{3}$ is one of
several due to Makai, and equality is approached by rectangles as they become
long and slender, leaving the question as to the best constant $c$ when
restricted to tangential polygons.

From [69] p254, for a thin rectangle

$Q_{0}\sim\frac{1}{3}\rho^{2}A\ \ {\rm for}\ \ \rho\rightarrow{0}.$

For thin isosceles triangles, with small vertex angle,

$Q_{0}\sim\frac{1}{6}\rho^{2}A\ \ {\rm for}\ \ \rho\rightarrow{0}.$

For the lower bound found in Part I

$Q_{0,-}\sim\frac{21}{128}\rho^{2}A\ \ {\rm for}\ \ \rho\rightarrow{0}.$

This checks with $Q_{0-}\leq Q_{0}$.

Another inequality for convex domains is

$\frac{1}{3}\frac{A^{3}}{L^{2}}\leq Q_{0}\leq\frac{2}{3}\frac{A^{3}}{L^{2}}.$

The left inequality is approached by thin rectangles. The right inequality is
approached by thin acute isosceles triangles. (It is curious that thin
rectangles occur as upper bounds in one inequality in this section, and lower
bounds in another.) However, inequalities for convex sets,

$\frac{\rho}{2}\leq\frac{A}{L}\leq\rho(1-\frac{\pi\rho}{L})\leq\rho,$

(in which the central inquality is equality for a disk, [80]) are consistent
with

$\frac{1}{3}\frac{A^{3}}{L^{2}}\leq Q_{0}\leq\frac{1}{3}\rho^{2}A.$

For tangential polygons the inequalities above give

$\frac{1}{2}\frac{A^{3}}{L^{2}}=\frac{1}{8}\rho^{2}A\leq
Q_{0}\leq\frac{1}{6}\rho^{2}A=\frac{2}{3}\,\frac{A^{3}}{L^{2}}.$

The extreme domains are the disk (left) and thin isosceles triangles (right).

See also [14].

###  22.2 Cheeger constant and $Q_{0}$

We have

$Q_{0}(\Omega)h(\Omega)^{4}\geq Q_{0}(B)h(\Omega)^{4}=2\pi.$

Equality occurs only for disks. See [57], and specialize to 2-dimensions. Thus
for our tangential polygons

$Q_{0}(\Omega)\geq\frac{32\pi A^{4}}{(L+\sqrt{4\pi A})^{4}}.$

Much is known about the elastic torsion function and, in particular, its
properties in convex domains.

It is known that in convex $\Omega$, the square root of the torsion function
$\sqrt{u_{0}}$ is concave. In some proofs of this one uses that for solutions
of the torsion equation $\log(1-4H)$, with H the hessian determinant, is
harmonic. $u_{0}$ is, itself, not concave in any domain with corners. However
one can ask on what subset $\Omega_{1}$ of $\Omega$ is $u_{0}$ concave. For an
equilateral triangle (and for a disk), $\Omega_{1}$ contains the incircle.
(See §11 of [45].) One wonders what additional conditions, if any, might be
needed for it to happen in other tangential polygons that $\Omega_{1}$
contains the incircle.

Improvements on the St Venant inequality involving Fraenkel asymmetry are
given in [11].

There is a Urysohn’s type inequality which we denote by (U).
(U): among convex sets with given mean width, the torsional rigidity is
maximized by balls.
The mean width $w$ of any compact shape $\Omega$ in two dimensions is $L/\pi$,
where $L$ is, as before, the perimeter of the convex hull of $\Omega$. So $w$
is the diameter of a circle with the same perimeter as the convex hull. So in
(U) it is just maximizing with same perimeter. Inequality (U) is weaker than
St Venant.

There are inequalities involving polar moment of inertia about the centroid.
For convex domains $Q_{0}I_{c}A^{-4}$ is maximized by its value for an
equilateral triangle. (Related results, if not exactly this, are in [68].)
We conjecture:
For tangential polygons $Q_{0}I_{2}A^{-4}$ is maximized by its value for an
equilateral triangle.

Sperb $u_{\rm max}\leq\rho^{2}$. Strip $u_{0}=(\rho^{2}-x^{2})/2$

##  23 Some bounds on $Q_{0}$

Recall the well-known estimates

$\displaystyle Q_{0}$ $\displaystyle\leq$
$\displaystyle\frac{\pi}{8}\,\left(\frac{A}{\pi}\right)^{2}=Q_{\odot,0},$
$\displaystyle Q_{0}$ $\displaystyle\leq$ $\displaystyle\Sigma_{\infty}.$

Define, in the notation of [69],

$B_{\Omega}=\int_{\partial\Omega}\frac{1}{x\cdot{n}}.$ (23.1)

Some well known lower bounds are:

$\displaystyle Q_{0}$ $\displaystyle\geq$ $\displaystyle\frac{A^{2}}{4\
B_{\Omega}}(={\rm{\ for\ tangential\ polygons\ }}\frac{A^{3}}{2L^{2}}),$
$\displaystyle Q_{0}$ $\displaystyle\geq$
$\displaystyle\frac{\pi}{8}\,{\dot{r}}^{2},$

where ${\dot{r}}$ is the maximum interior mapping radius of $\Omega$.

##  24 More on isosceles triangles

###  24.1 Isoperimetric inequalities for isosceles triangles

Numerical data on functionals associated with isosceles triangles was given
in§9. Our main emphasis in the following will be isosceles triangles rather
than general triangles. However, some general statements are available. Recall
that $\cal A$ is all (simply-connected) domains, ${\cal A}_{n}$ is all
$n$-gons.

  * •

$Q_{0}/A^{2}$ is maximized over ${\cal A}_{3}$ (triangles) for an equilateral
triangle.

  * •

$B=L/\rho$ is minimized over ${\cal A}_{3}$ (triangles) for an equilateral
triangle.

  * •

${\dot{r}}/\sqrt{A}$ is maximized over ${\cal A}_{3}$ (triangles) for an
equilateral triangle.

####  24.1.1 $B$

$B=L/\rho$ for any tangential polygon, so for isosceles triangles in
particular.
The disk minimizes $B$ and $Q_{0}B/A^{2}$ over tangential polygons.
The equilateral triangle minimizes $B$ and $Q_{0}B/A^{2}$ over triangles.

shape | $n$ | $8\pi Q_{0}/A^{2}$ | $Q_{0}/(A/\pi)^{2}$ | $B=\frac{L}{\rho}$ | $4Q_{0}B/A^{2}$
---|---|---|---|---|---
disk | $\infty$ | 1 | $\pi/8$ | $2\pi$ | $1$
hexagon | 6 | 0.9643 | 0.3786 | $4\sqrt{3}$ | $1.063$
square | 4 | 0.8834 | 0.3469 | $8$ | $1.125$
equilateral $\Delta$ | 3 | 0.7255 | 0.2849 | $6\sqrt{3}$ | $1.200$
right isosceles |  | 0.6557 | 0.2575 | $4(1+\sqrt{2})$ | $1.217$

####  24.1.2 $\dot{r}$

The calculation of $\dot{r}$ for polygons often involves the use of Schwarz-
Christoffel conformal mappings. Some exact values are given on p273 of [69].

  * •

(Aside, except for $n=3$.) For a regular polygon with $n$ sides, and perimeter
$L_{n}$,

${\dot{r}}_{n}=\frac{\Gamma(1-\frac{1}{n})}{2^{1+2/n}\,\Gamma(\frac{1}{2})\,\Gamma(\frac{1}{2}+\frac{1}{n})}\,L_{n}.$

  * •

Again citing [69] p158, amongst all triangles with a given area that which
maximizes $\dot{r}$ is equilateral.

  * •

For (regular polygons and) triangles we have

$\pi{\dot{r}}{\bar{r}}=A,$

$A$ being the area and $\bar{r}$ the transfinite diameter.

  * •

Working from earlier more general results (Haegi 1951) in [22] it is given
that for an isosceles triangle, vertex angle $\alpha$ and base
$2\sin(\alpha/2)$ the transfinite diameter, denoted there by $\kappa$, is

$\kappa(\alpha)=\frac{\sqrt{\pi+\alpha}}{8\pi^{5/2}}\left(\frac{\pi+\alpha}{4\alpha}\right)^{\alpha/(2\pi)}\,\frac{\sin(\alpha)^{2}}{\sin(\alpha/2)}\,\Gamma(\frac{\alpha}{\pi})\,\Gamma(\frac{\pi-\alpha}{2\pi})^{2}.$

For isosceles triangles with area $A$ and vertex angle $\alpha$

${\bar{r}}=\frac{2\sqrt{A\tan(\alpha/2)}}{\sin(\alpha/2)}\,\kappa\ \ {\rm so\
\
}{\dot{r}}=\frac{A}{\pi{\bar{r}}}=\sqrt{A}\,\frac{\sin(\alpha/2)}{2\pi\sqrt{\tan(\alpha/2)}}\,\frac{1}{\kappa}.$

Some values of $\dot{r}$, as given in [69], are copied in the following table.

shape |  | $\dot{r}$ | $\dot{r}$ | $\frac{\dot{r}}{\sqrt{A}}$ | $8Q_{0}/(\pi{\dot{r}}^{4})$
---|---|---|---|---|---
disk | radius $a$ | $a$ | $a$ | $0.56419$ | $1$
hexagon | side $s_{6}$ | $\frac{2^{5/3}\,\sqrt{3}\pi}{\Gamma(1/3)^{3}}\,s_{6}$ | $0.89850s_{6}$ | $0.55744$ | $1.011$
square | side $s_{4}$ | $\frac{4\sqrt{\pi}}{\Gamma(1/4)^{2}}\,s_{4}$ | $0.53835s_{4}$ | $0.53935$ | $1.058$
equilateral $\Delta$ | side $s_{3}$ | $\frac{2\pi}{\Gamma(1/3)^{3}}\,s_{3}$ | $0.3268\,s_{3}$ | $0.49665$ | $1.209$
right isosceles | equal sides $a$ | $\frac{4\sqrt{2\pi}}{3^{3/4}\,\Gamma(1/4)^{2}}\,a$ | $0.33462a$ | $0.47320$ | $1.325$

####  24.1.3 Simple comments on torsion for isosceles triangles

As noted before, for an equilateral triangle the torsion function is a cubic
polynomial in $x$ and $y$ corresponding to forming the products of three
linear terms, each linear term being 0 on one side of the triangle. There
appear to be no other simple solutions for isosceles triangles, though there
is a series formula for the torsional rigidity of the right isosceles
triangle.

In the table below, in the second, third and fourth columns we take the base
to be 2, the area to be $h$, $h$ being the height; in the fifth and sixth
columns the area is $\sqrt{3}$, the height is $h$ and the base $2\sqrt{3}/h$.

infinitely acute | $h\rightarrow\infty$ | $A\sim{h}$ | $Q_{0}\sim\frac{1}{6}h$ | $h\rightarrow\infty$ | $Q_{0}\sim\frac{1}{2h}$
---|---|---|---|---|---
equilateral | $h=\sqrt{3}$ | $A=\sqrt{3}$ | $Q_{0}=\frac{\sqrt{3}}{20}$ | $h=\sqrt{3}$ | $Q_{0}=\frac{\sqrt{3}}{20}$
right isosceles | $h=1$ | $A=1$ | $Q_{0}=0.026091$ | $h=3^{1/4}$ |
infinitely flat | $h\rightarrow{0}$ | $A\sim{h}$ | , $Q_{0}\sim\frac{1}{24}h^{3}$ | $h\rightarrow{0}$ | $Q_{0}\sim\frac{h}{8}$

####  24.1.4 More variational bounds on $Q_{0}$ for isosceles triangles

The bounds on $Q_{0}$ we report here are often very old.

There are, of course, variational formulations of the Problem (P(0)). For
positive, differentiable functions $v$, let

$E(v,0,k)=\int_{\Omega}v^{k}\qquad{\rm and\ \ }E(v,1,k)=\int_{\Omega}|\nabla
v|^{k}\ .$

With

$Q_{0,\rm LB}(v)=E(v,0,1)^{2}/E(v,1,2)$ (24.1)

it can be shown that for any smooth function $v$ vanishing on the boundary of
$\Omega$, the expression $Q_{0,\rm LB}(v)$ provides a lower bound on the
torsional rigidity $Q_{0}=Q_{0,\rm LB}(u)$. The theory for this is given in
[69].

For isosceles triangles, base $2a$, height $h$, a simple trial function $v$ to
substitute into (24.1) is

$v_{\rm cub}=(y+\rho)(1-(x/a)-((y+\rho)/h))(1+(x/a)-((y+\rho)/h)).$ (24.2)

Evaluating the integrals gives

$Q_{0\rm{cub}}=Q_{0,\rm LB}(v_{\rm
cub})=\frac{a^{3}h^{3}}{(30a^{2}+10h^{2})}=\frac{A^{2}}{30\tau+10/\tau},$
(24.3)

with $\tau=a/h=\tan(\alpha/2)$ as before, and the expression on the right of
(24.3) is, in fact, an approximation given in Roark’s tables, valid for the
the vertex angle $\alpha$ of the isosceles triangle lying between 40 and 80
degrees. It is exact for the equilateral case where the vertex angle is 60
degrees. It would, we think, be an improvement to Roark’s tables to have
noted, after defining this rational function of $a$ and $h$, to have noted
that $Q_{0}>Q_{0,\rm LB}(v_{\rm cub})$ for all positive values of $a$ and $h$,
and after this to have noted the range of $h/a$ over which it provides a good
approximation to $Q_{0}$.

We can compare this lower bound with our earlier $Q_{0-}$ as shown in Figure
1. In the next figures the bound of (24.3) is shown brown, dashed.
Unsurprisingly it is good for triangles near equilateral improving on $Q_{0-}$
there, but it is worse than $Q_{0-}$ when not near equilateral. For example,
for a right isosceles triangle, with area $\sqrt{3}$ we found

$Q_{0}=0.07827,\qquad Q_{0-}=0.07651,\qquad Q_{0\rm{cub}}=\frac{3}{40}=0.075.$

Figure 14: 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 $Q_{\Delta}$, brown dashed is the cubic one underdiscussion.

Figure 15: 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 $Q_{\Delta}$, brown dashed is the cubic one underdiscussion.

Define $v_{\rm quad}$ to be even in $x$ and

$v_{\rm quad}=y(1+(x/a)-(y/h))\ \ {\rm for}\ \ 0<x<a.$

Good lower bounds on $Q_{0}$ for when $h$ is small can be found using this as
test function.

See PartI, near Figure 2 for another lower bound on $Q_{0}$. Many other bounds
are available. See for example the Appendix, by Helfenstein of the paper [67].

ToDo. Check the Helfenstein work. Stretching in one direction Define

$D(1,t)=\\{(tx,y)|(x,y)\in D\\}.$

In particular, $D(1,1)=D$. Domain monotonicity gives $QQ_{0}(D(1,t))$ is
increasing in $t$. In [67] it is shown that $t/Q_{0}(D(1,t))$ is increasing
and concave in $t^{2}$. The appendix to their paper by Helfenstein makes good
use of this in connection with isosceles triangles, finding upper and lower
bounds on $Q_{0}$ which differ by no more than 12%. These are, unfortunately,
rather elaborate, and with cheap numerical computing, it is perhaps better to
use the stretching result as yet another check on numerics.

ToDo. Main interest is in $Q_{0}(D(1,t))/|D(1,t)|^{2}$ where the maximum
occurs for $t$ giving an equilateral triangle.

###  24.2 Miscellaneous other bounds

Consider an isosceles triangle with a base of length 2 and height of length
$h$, and angles $\beta$, $\beta$, and $\pi-2\beta$ respectively. Since
$\tan(\beta)=h$, and the area $A$ of the triangle is $h$, it can be shown that
an inequality from [9] is

$Q_{0}\leq\frac{1}{8}(1+A^{2})^{2}\,(A-{\rm arctan}(A)).$

This inequality is, in [9], used with triangles which are thin, and, while
satisfied for equilateral, it is weak there.

##  25 General triangles

From the calculations of $L$, $i_{2}$ nad $i_{4}$ in Part IIb §16,
$\Sigma_{\infty}$ and $\Sigma_{1}$ can be found.

##  26 Rhombi

The formulae for $A$, $L$, $i_{2}$, $i_{4}$, $\Sigma_{\infty}$, $\Sigma_{1}$
are given in Part IIa §10. The numerical values of $Q_{0}$ are from [79] (and
checked with Mathematica NDSolve).

Figure 16:  Rhombus

##  27 Blaschke-Santalo for $Q_{0}$

If one were to assemble, for tangential polygons with given $\rho$ and
$d_{O}$, a Blaschke-Santalo diagram, I would expect that with $Q_{0}$ would be
similar to that with the area (or equivalently the perimeter) given earlier.
In particular, at given $\rho$ and $d_{O}$,

  * •

the 1-cap would minimize $Q_{0}$ thereby providing the left-upper boundary in
the diagram, and

  * •

the regular polygons would be amongst the maximizing shapes thereby providing
a scatter of points on the right-lower boundary.

There may be independent interest in curves, or other subsets, in the region,
e.g. for

  * •

triangles

  * •

kites/rhombi
other quadrilaterals.

A much easier task is to find the diagram for the lower bound $Q_{0-}$. I
expect everything stated above for $Q_{0}$ would occur in this simpler
situation.

##  28 Further questions

Question. Does ‘corner cutting’ (defined in §12) increase $Q_{0}/A^{2}$?
It would be (much) easier to begin with $Q_{0-}$ and isosceles triangles each
having its apex corner cut to form a symmetric tangential trapezium.

Given the area of a regular $n$-gon, its inradius is, as in Part IIa §8,

$\rho_{n}^{2}=\frac{A}{n\tau_{n}}\qquad{\rm with\ \
}\tau_{n}=\tan(\frac{\pi}{n}).$

Formulae for $i_{2,n}$, $i_{4,n}$, $\Sigma_{1,n}$ and $\Sigma_{\infty,n}$ are
also given in Part IIa §8. Let $f_{n}(Q_{0})$ be the quadratic in $Q_{0}$
defined at Part I, equation (3.6) using the values appropriate to the regular
$n$-gon for $\rho$, $i_{2}$ and $i_{4}$. The leading coefficient, i.e.
coefficient of $Q_{0}^{2}$, is the same for the two polynomials.

For an equilateral triangle with area $\sqrt{3}$, we found in Part I

$f_{3}(Q_{0})=f_{\Delta}(Q_{0})=32\sqrt{3}(Q_{0-}\frac{\sqrt{3}}{20})(Q_{0}-\frac{\sqrt{3}}{4}).$

We wish to show that, for any tangential $n$-gon with the same area

$Q_{0-}\leq Q_{0-,n}.$

This suggests that we try to show

$f(Q_{0,n})\leq{0},\qquad f_{n}(Q_{0})\geq{0}.$

This leads to the question of how roots of a quadratic change when its
coefficients change. Both polynomials $f$ are of the form

$32A(Q_{0}^{2}+b*Q_{0}+c)=0,$

with two positive roots

$2\sqrt{c}<b<0,\qquad 0<c<\frac{b^{2}}{4}.$

If we could show that both coefficients $b$ and $c$ are largest for the
regular $n$-gon case, this would establish that $Q_{0,n}$ was larger than any
other tangential $n$-gon.

It would seem prudent

  * •

to begin with $n=3$,

  * •

then $n=4$, initially with special cases (rhombi, kites, etc.)

  * •

then $n=5$,

  * •

then $n=6$,

  * •

then general $n$.

Part IV: Robin boundary conditions $Q(\beta)$,
isoperimetric inequalities, etc.

##  29 Robin boundary conditions

Part IV concerns a different problem than the earlier parts.

The function $R$ of [47] is a function of a nonnegative parameter $\beta$ and
geometric functionals $A$, $L$, $\Sigma_{\infty}$ and $\Sigma_{1}$ and
$Q_{0}$. Knowledge of these, for triangles, (as in Part II) is essential for
the following questions.

Let $\beta\geq{0}$. For solutions $u_{\beta}$ of Problem (P($\beta))$ of [47]
in a triangle define $Q_{\rm triangle}(\beta)$ as the integral of $u_{\beta}$
over the triangle.

Question. Consider triangles of fixed area, say $\sqrt{3}$. Amongst these
triangles is that which has the greatest $Q(\beta)$ the equilateral triangle?
(For $\beta=0$ this is the case as proved in [69].) (A similar question, but
for the fundamental frequency is asked in [55].)

In [47] we presented a lower bound $R(\beta,\ldots)$ for $Q(\beta)$ and noted
that it provided a good approximation to $Q(\beta)$. For triangles the only
argument of $R$ that isn’t a relatively simple function of the triangle’s
geometry is the torsional rigidity $Q_{0}$. This leads on to simpler questions
(for which some of Part III might be relevant).

Question. Consider triangles of fixed area, say $\sqrt{3}$. What additional
properties of the torsional rigidity $Q_{\rm triangle}(0)$ will ensure that
for these triangles that which has the greatest $R(\beta,\ldots)$ is the
equilateral triangle?

Questions. Consider triangles of fixed area, say $\sqrt{3}$. Replacing, in
$R(\beta,\ldots)$, $Q_{\rm triangle}(0)$ by one of its bounds or
approximations, will ensure that for these triangles that which has the
greatest $R(\beta,\ldots,Q_{0,{\rm approx}})$ is the equilateral triangle?
(The plural in ‘Questions’ is because there are several possible bounds that
might be used.)

There is some indication via numerics that, restricting to isosceles
triangles, equilateral triangles maximize.

More generally, similar questions can be asked for $n$-gons. For these, does
the regular $n$-gon optimize?

Part V: Miscellaneous PDE and tangential polygons

Tangential polygons get a mention in other pde papers.

Papers by Solynin include treatment of the fundamental frequency for
tangential $n$-gons.

Consider evolution of temperature satisfying the heat equation, zero Dirichlet
data on boundary of a convex plane domain $C$, and unit initial value. In the
convex domain $C$ there will, at each fixed time, be a unique point at which
the solution is a maximum, hot-spot. [58] characterises the tangential
polygons which have stationary hot-spots, basically as the regular polygons.
We remark that the integral of the solution to the heat equation, integrated
w.r.t. time $t$ from 0 to infinity, solves the torsion problem. So the unique
maximum of the torsion function would have to be the stationary hot-spot. They
must coincide with the unique critical point of the first eigenfunction. What
further restrictions are needed on a tangential polygon in order that the
incentre is a stationary hot spot?

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