A Study of Certain Sharp Poincaré Constants as Set Functions of Their Domain

Abstract

For bounded, convex sets $\Omega \subset \mathbb {R}^d$, the sharp Poincaré constant $C(\Omega )$, which appears in $||f-\bar{f}_{_{\Omega }}||_{L^{\infty }(\Omega )} \le C(\Omega )||\nabla f||_{L^{\infty }(\Omega )}$, is given by $C(\Omega )=\max _{_{\partial \Omega }}\zeta$ for a specific convex function $\zeta$ [Bevan et al. in Proc Am Math Soc 151:1071–1085, 2023 (Theorem 1.1)]. We study $C(\cdot )$ as a function on convex sets, in particular on polyhedra, and find that while a geometric characterization of $C(\Omega )$ for triangles is possible, for other polyhedra the problem of ordering $\zeta (V_i)$, where $V_i$ are the vertices of $\Omega$, can be formidable. In these cases, we develop estimates of $C(\Omega )$ from above and below in terms of more tractable quantities. We find, for example, that a good proxy for C(Q) when Q is a planar polygon with vertices $V_i$ and centroid $\gamma (Q)$ is the quantity $D(Q)=\max _{i}|V_i-\gamma (Q)|$, with an error of up to $\sim 8\%$. A numerical study suggests that a similar statement holds for k-gons, this time with a maximal error across all k-gons of $\sim 13\%$. We explore the question of whether there is, for each $\Omega$, at least one point M capable of ordering the $\zeta (V_i)$ according to the ordering of the $|V_i-M|$. For triangles, M always exists; for quadrilaterals, M seems always to exist; for 5-gons and beyond, they seem not to.

1 Introduction

For a general bounded, convex set $\Omega \subset \mathbb {R}^d$ with $\mathcal {L}^d(\Omega )>0$, let

(1.1)

and define

$$\begin{aligned} C(\Omega ):=\max _{x\in \partial \Omega } \zeta (x;\Omega ). \end{aligned}$$

(1.2)

It has recently been shown in [1, Theorem 1.1] that $C(\Omega )$ is the sharp Poincaré constant in the inequality

$$\begin{aligned} ||f-\bar{f}_\Omega ||_{L^{\infty }(\Omega )} \le C(\Omega )||\nabla f||_{L^{\infty }(\Omega )} \quad \quad f \in W^{1,\infty }(\Omega ,\mathbb {R}^d), \end{aligned}$$

where . This paper aims to understand the behaviour of $C(\Omega )$ as a function on convex sets. The function $\zeta$ is itself convex independently of $\Omega$, which makes it particularly attractive to study when $\Omega$ is a polyhedral set. Indeed, Bauer’s maximum principle [2, 3] immediately implies that $\zeta$ necessarily attains its maximum over $\overline{\Omega }$ at one (or more) of its vertices. We therefore focus, in this work, on polyhedral sets, and our aim is to deduce, where possible, geometric criteria that deliver $\max _{\partial \Omega }\zeta$. This turns out to be relatively straightforward when $\Omega$ is a triangle, but surprisingly difficult otherwise, even for planar polyhedra (i.e. polygons) and despite an explicit formula for $\zeta (V_i;\Omega )$ always being available in this case. Using the latter, together with a suitable computer code or package, one can compute directly $\zeta (V_1;\Omega ), \ldots , \zeta (V_k;\Omega )$ for $\Omega$ a k-gon with vertices $V_1,\ldots ,V_k$, and so determine the maximum. When carried out on a variety of polygons, these calculations can hint at the underlying geometric principles that we are most interested in finding. They are, however, far from being our only source of ideas.

In this paper, we use several approaches to the problem outlined above. In the case of a triangle, direct proofs using (1.1) together with (a) a layer-cake representation for $\zeta (V;\Omega )$, see Proposition 2.1, and (b) a gradient flow argument, see Proposition 2.4, lead to the characterization that $\max \zeta (\cdot ;\Omega )$ occurs at the vertex with the smallest interior angle, which accords with intuition. For triangles and more general polyhedra, we introduce in the ideas of the singly isolated vertex (SIV), see Definition 3.2, and internally reflected region (IRR), see Definition 3.4, a means of comparing $\zeta (V_i;\Omega )$ with $\zeta (V_j;\Omega )$. The argument is given in Lemma 3.6. When one can do this for all i, j in $1,\ldots ,k$, which is sometimes the case, $\max \zeta (\cdot ;\Omega )$ can be determined.

In Sects. 3.3 and 3.4 the emphasis changes to finding upper and lower bounds on $C(\Omega )$. For example, in the case of a planar, bounded, convex quadrilateral Q, Proposition 3.13 provides the sharp estimate

$$\begin{aligned} \frac{1}{4} \textrm{codiam}\,(Q) \le \max _{\partial Q}\zeta (\cdot \,;Q) \le \frac{2}{3} \textrm{diam}\,(Q). \end{aligned}$$

The quantity $\textrm{codiam}\,(\Omega )$ is given in general by (3.14) and, in the case of the quadrilateral, by $\textrm{codiam}\,(Q)=|V_1-V_3|+|V_2-V_4|$. The theme of bounding $C(\Omega )$ from above and below is carried further in Proposition 3.17, where it is shown that the maximum distance of the centroid from the vertices $V_i$ of Q acts as a proxy for $\max _{\partial Q} \zeta (\cdot ;Q)$ in the sense that

$$\begin{aligned} \max _i|\gamma (Q)-V_i| \le \max _{\partial Q} \zeta (\cdot \,;Q) \le r_u \max _i|\gamma (Q)-V_i|.\end{aligned}$$

(1.3)

A numerical investigation suggests that $r_u =r_u(\text {reg})=(2-\sqrt{2}\ln (\sqrt{2}-1))/3$, which, being approximately 1.0822, means that there is about an $8\%$ error in calculating $\max _{\partial Q} \zeta (\cdot ;Q)$ in this way. The quantity $r_u(\text {reg})$ corresponds to taking Q to be the regular 4-gon, i.e. a square, which choice ‘saturates’ the upper bound. We also find that estimates of the form (1.3) appear to hold for all planar k-gons, with the prefactors $r_u$ increasing monotonically with k to $\frac{32}{9\pi }$, which happens to be the Poincaré constant C(B(0, 1)) for the disk of unit radius. It is natural to wonder whether we can replace the centroid $\gamma (Q)$ in (1.3) with some other point derived from Q in such a way that the ‘percentage error’ decreases. This leads to the idea of ‘magic points’ (see (3.38) and (3.39)), which we discuss in Subsect. 3.5, as a means of ordering the values of $\zeta (V_i;\Omega )$ for $i=1,\ldots , k$.

There are a number of sharp Poincaré constants available in the literature, a useful guide to which is the article [4], or else see [5] and the references therein. Significant among these are the classical results of Szego [6], Weinberger [7], and Payne and Weinberger [8] who, for a convex domain G, characterize the sharp Poincaré constant in terms of

$$\begin{aligned} \mu _2(G)=\inf _{\bar{u}_{_{G}}=0} \frac{\int _G |\nabla u|^2 \, dx}{\int _G u^2 \, dx}. \end{aligned}$$

We are interested in the extent to which such quantities can be made explicit. This is the case, for instance, in the sharp inequality

$$\begin{aligned} ||u||_{L^1(\Omega )} \le \frac{\textrm{diam}\,(\Omega )}{2} ||\nabla u||_{L^1(\Omega )} \quad \quad u \in W^{1,1}(\Omega ) \end{aligned}$$

given by Acosta and Durán in [9] (with $\bar{u}_{\Omega }=0$), and in the series of sharp inequalities given by Cianchi in [10, Theorem 3]. The characterization given in [1] (see (1.2) above) is not quite in this class: there is, for general convex $\Omega$, more work to be done, which is one reason we investigate the properties of $\max _{\partial \Omega }\zeta (\cdot ; \Omega )$ in this paper.

It may also be interesting to note that $\zeta (x)$ and $C(\Omega )$ have a certain mechanical interpretation too, which are, respectively, that $\zeta (x)$ represents the moment of mass about x of a uniformly dense material occupying the domain $\Omega$, and $C(\Omega )$ measures the maximum such moment over points in $\overline{\Omega }$. Perhaps there is some deeper significance to this observation.

We conclude this introduction by noting that, for general bounded, open and convex domains $\Omega \subset \mathbb {R}^d$, a direct calculation gives

from which we deduce that $\zeta$ is subharmonic. It follows from the maximum principle (see, for example, [11, Theorem 2.2]) that if the maximum of $\zeta$ occurs at an interior point of $\Omega$ then $\zeta$ is constant. But when $\Omega$ is such that its d-dimensional (Lebesgue) measure $\mathcal {L}^d(\Omega )>0$, $\zeta$ is strictly convex (see [1, Eq. (2.3) and ensuing calculation]), and in particular $\zeta$ is not constant. Hence, $\max _{\overline{\Omega }}\zeta$ is attained on $\partial \Omega$, which is a partial recovery of the result of Bauer’s [2, 3] that the maximum of $\zeta$ should be attained at an extreme point of $\Omega$. In fact, applying the strict convexity of $\zeta$ recovers the result in full.

1.1 Notation and Plan of the Paper

We write for $\int _{\Omega } f(x)\, d\mathcal {L}^d(x) /\mathcal {L}^d(\Omega )$. Any other notation that we use is, unless standard, explained at the point of introduction.

The plan of the paper has already been hinted at in the introduction. To be more precise, in Sect. 2 we describe in detail the characterization of C(T) when T is a triangle. Section 3 studies $C(\Omega )$ for more general polygons and includes, in Subsects. 3.3 and 3.4, a development of quantitative upper and lower bounds for $C(\Omega )$, and, in Subsect. 3.5, results concerning the existence (and non-existence) of ‘magic points’ for various planar polygons.

2 A Characterization of $\max _T \zeta (\cdot ;T)$ When T is a Triangle

Let $\Omega$ be any measurable subset in $\mathbb {R}^d$ such that $0<\mathcal {L}^d(\Omega )<\infty$. Then, by [12, Lemma 1.5.1] we employ the layer cake representation

$$\begin{aligned} \zeta (x;\Omega )=\frac{1}{\mathcal {L}^d(\Omega )} \int _0^\infty \mathcal {L}^d(\{y \in \Omega : \ |x-y|>t\})\, d\mathcal {L}^1(t). \end{aligned}$$

(2.1)

That is, $\zeta (x):=\zeta (x;\Omega )$ can equivalently be obtained by integrating (with respect to t) the volume fraction of $\Omega$ that lies outside the ball B(x, t). To illustrate its use, we prove the following intuitively clear result concerning $\max _{T}\zeta$, where T is a triangle.

Proposition 2.1

Let T be a triangle with vertices $V_1,V_2$ and $V_3$ and corresponding angles $\alpha _1$, $\alpha _2$ and $\alpha _3$ labelled so that $\alpha _1 \le \alpha _2 \le \alpha _3$. Then $\max _{x \in T} \zeta (x) = \zeta (V_1)$.

Proof

First note that there is no loss of generality in supposing that T has been labelled as described, and by [2, 3] the maximum of $\zeta$ will occur at one of the vertices. The result is trivial if T is equilateral, so we can suppose that $\alpha _1 < \pi /3$. For each vertex V and $t>0$ we define the ‘excluded volume fraction’ function

$$\begin{aligned} \nu (V,t)= \frac{\mathcal {L}^2(\{y \in T: \ |V-y|>t\})}{\mathcal {L}^2(T)}. \end{aligned}$$

Note that the numerator is just $\mathcal {L}^2(T\setminus B(V,t))$. Now, for sufficiently small t, it is clear that the excluded volume fractions are ordered according to

$$\begin{aligned} \nu (V_1,t) \ge \nu (V_2,t) \ge \nu (V_3,t),\end{aligned}$$

(2.2)

which follows directly from the ordering $\alpha _1\le \alpha _2 \le \alpha _3$ and the fact that for small enough t the sets $B(V_j,t)\cap T$ are sectors of increasing area as j runs from 1 to 3. The force of the rest of the proof is that the trend established in (2.2) continues to hold, even when the sets $B(V_j,t) \cap T$ cease to be sectors. We can suppose without loss of generality that T is as shown in Fig. 1 below, and we focus first on showing that

$$\begin{aligned} \zeta (V_2)\ge \zeta (V_3).\end{aligned}$$

(2.3)

Fig. 1

For illustration, $B(V_3,t)$ excludes a greater proportion of the area of T than $B(V_2,t)$ because ‘more’ of T lies in the same half-space as $V_3$ than lies in the same half-space as $V_2$

Let $L_{23}$ be the perpendicular bisector of the edge $V_2V_3$ and let $d_{23}$ be the length of the edge $V_2V_3$ and $d_{13}$ that of $V_1V_3$. Since $\alpha _1 \le \alpha _2$, we have $d_{23}\le d_{13}$, and since the ordering (2.2) clearly holds for $t\le \frac{d_{23}}{2}\le \frac{d_{13}}{2}$, we may as well suppose that $t > \frac{d_{23}}{2}$. In this case, the boundaries of the two disks $B(V_2,t)$ and $B(V_3,t)$ meet in two points on $L_{23}$, and, because (a) $\alpha _3 \ge \alpha _2$ and (b) the positions of the disks $B(V_2,t)$ and $B(V_3,t)$ are symmetric relative to $L_{23}$, it follows that $B(V_3,t)$ excludes a greater proportion of the area of T than does $B(V_2,t)$. But this says precisely that $\nu (V_2,t)\ge \nu (V_3,t)$ when $t>\frac{d_{23}}{2}$, and since we have already established, in (2.2), that the same inequality holds when $t \le \frac{d_{23}}{2}$, integrating with respect to t and applying (2.1) gives (2.3). Hence $\max _{x \in T}\zeta (x)$ must be one of $\zeta (V_1)$ and $\zeta (V_2)$. But, by considering $L_{12}$, the perpendicular bisector of the edge $V_1V_2$ and by arguing as above (this time making use of the fact that $\alpha _2\ge \alpha _1$), we see that for any $t >\frac{d_{12}}{2}$ it must be that $\nu (V_1,t)\ge \nu (V_2,t)$. Integrating as above and applying (2.1) gives $\zeta (V_1) \ge \zeta (V_2)$, and hence the conclusion of the proposition. $\square$

The argument above is a geometric version of the following, apparently much less straightforward, analytical one. Borrowing from (3.5) for now, we can write down explicit expressions for $\zeta (V_1)$ and $\zeta (V_2)$ for any triangle T, and then compare the two.

Proposition 2.2

Let T be a triangle with vertices $V_1$, $V_2$ and $V_3$, let $\alpha _1$, $\alpha _2$ and $\alpha _3$ be the corresponding angles, and let $\delta =|V_1V_2|$. Then

$$\begin{aligned} \zeta (V_1)&= \frac{\delta ^3}{6|T|}\sin ^3(\alpha _2)(h(\alpha _1+ \alpha _2)-h(\alpha _2)) \\ \zeta (V_2)&= \frac{\delta ^3}{6|T|}\sin ^3(\alpha _1)(h(\alpha _1+ \alpha _2)-h(\alpha _1)), \end{aligned}$$

where $h:(0,\pi ) \rightarrow \mathbb {R}$ is given by (3.2). In particular, $\zeta (V_1) \ge \zeta (V_2)$ if and only if

$$\begin{aligned} \sin ^3(\alpha _2) f(\alpha _1,\alpha _2) \ge \sin ^3(\alpha _1) f(\alpha _2,\alpha _1), \end{aligned}$$

(2.4)

where $f(\alpha ,\beta ):=h(\alpha +\beta ) -h(\beta )$.

Proof

This follows by a direct calculation from (3.3). $\square$

Our goal is to establish, by analytical means, that (2.4) holds if and only if $\alpha _1\le \alpha _2$. This is the statement of Proposition 2.4 below, whose proof relies on the following technical lemma.

Lemma 2.3

Let the set E be given by

$$\begin{aligned} E:=\{(a,b): \ 0< a< \min \{b,\pi -b\}, 0< b < \pi \}\end{aligned}$$

(2.5)

and define the vector field $X=(X_1,X_2)^T$ on E by

$$\begin{aligned} X(A) = \left( \begin{array}{l} 4+\rho (A)-3\sin ^2 \alpha _1 \cos \alpha _1 f(\alpha _2,\alpha _1) \\ 3\sin ^2 \alpha _2 \cos \alpha _2 f(\alpha _1,\alpha _2) + \rho (A) \end{array}\right) , \end{aligned}$$

(2.6)

where

$$\begin{aligned} \rho (A) = 2\left( \frac{\sin ^3 \alpha _2 - \sin ^3 \alpha _1}{\sin ^3(\alpha _1+\alpha _2)}-1\right) \end{aligned}$$

and $A=(\alpha _1, \alpha _2)^T$. Let $O=(0,0)$, $B=(0,\pi )$ and $C=(\pi /2,\pi /2)$ be the vertices of E. Then X has a natural extension to the lines OB and OC, excluding endpoints, such that

• (i) $X(0,b)=(1,0)^T$ if $0< b < \pi$;

• (ii) $X(a,a)=\mu (a)(-1,1)^T$ for $0< a < \frac{\pi }{2}$, where $\mu (a):=3\sin ^2a\cos a \,f(a,a) - 2 >0$;

• (iii) $\lim _{\epsilon \rightarrow 0} \frac{X_2(a,\pi -a-\epsilon )}{X_1(a,\pi -a-\epsilon )} = 1$, and

• (iv) $X_2(A) > 0$ if $A \in E$.

Proof

(i) By inspection, $X_2(0,b)$ is well defined and obeys $X_2(0,b)=0$ for $0< b < \pi$. For the same range of b, $4+\rho (0,b)=4$. $X_1(a,b)$ has a removable singularity at $a=0$ by observing that for fixed b, $3 \sin ^2 a \cos a \, h(a+b) \rightarrow 0$ as $a \rightarrow 0$, while

$$\begin{aligned} \lim _{a \rightarrow 0} -3 \sin ^2 a \cos a \, h(a) = 3 \end{aligned}$$

by referring to the definition of h(a). Hence $X_1(a,b) = 4+ \rho (a,b) - 3 \sin ^2 a \cos a (h(a+b) -h(a)) \rightarrow 1$ as $a \rightarrow 0$. This proves (i).

(ii) For $0< a < \pi /2$, it is clear that X(a, a) is both well defined and obeys $X(a,a)=\mu (a)(-1,1)^T$, with $\mu (a)$ as stated above. This function, when expressed in terms of $x:=\tan ^2(a/2) \in (0,1)$, takes the form

$$\begin{aligned} \mu (a)=\tilde{\mu }(x) = \frac{48 x (1-x)^2 \ln (2/(1-x)) + 1 - 28 x + 30 x^2 + 52 x^3 - 7 x^4 }{4(1-x)(1+x)^3}. \end{aligned}$$

For $x \in (0,1)$, the numerator can be written as

$$\begin{aligned} 1+(48 \ln 2-28)x + (78-96 \ln 2)x^2+(48 \ln 2 - 20) x^3 \\ + 9 x^4 + 96 \sum _{j=5}^{\infty } \frac{x^j}{(j-3)(j-2)(j-1)}, \end{aligned}$$

where each coefficient is positive, and the denominator is obviously positive for this range of x. Hence $\mu (a)>0$, as claimed.

(iii) Let $a \in (0,\pi /2)$. A short calculation shows that $X_1(a,\pi -a-\epsilon )$ and $X_2(a,\pi -a-\epsilon )$ are both equal to $\frac{3 \sin ^2 a \cos a}{\epsilon ^2} + o(\epsilon ^{-2})$ as $\epsilon \rightarrow 0$. Hence part (iii) of the statement of the lemma.

(iv) Let $A=(a,b)$ lie in E. A direct calculation using the definition of f(a, b), together with the fact that $h'(a)=2\csc ^3 a$, shows that

$$\begin{aligned}&\partial _a X_2(A) = \frac{6}{\sin ^4(a+b)}\\&\qquad \left( (\sin ^2 b \cos b - \sin ^2 a \cos a)\sin (a+b) - (\sin ^3 b - \sin ^3 a)\cos (a+b)\right) , \end{aligned}$$

which simplifies to

$$\begin{aligned} \partial _a X_2(A) = \frac{6 (\sin b - \sin a) \sin a \sin b}{\sin ^4(a+b)}.\end{aligned}$$

Let $b \in (0,\pi /2]$. Then in order for (a, b) to lie in E it must be that $a<b$, whence $\partial _a X_2(A) >0$. Similarly, if $b \in (\pi /2,\pi )$ and $(a,b) \in E$ then necessarily $0< a < \pi -b$, and hence $\sin a < \sin b$. Thus $\partial _a X_2(A) >0$ for all $A \in E$. To conclude the proof of part (iv), use part (i) of the lemma to note that $X_2(0,b)=0$, so that $X_2(A)=\int _0^a \partial _s X_2(s,b) \, ds >0$. $\square$

Proposition 2.4

Under the assumptions of Proposition 2.2, $\sin ^3(\alpha _2) f(\alpha _1,\alpha _2) \ge \sin ^3(\alpha _1) f(\alpha _2,\alpha _1)$ if and only if $\alpha _2 \ge \alpha _1$.

Proof

For clarity, let $a=\alpha _1$ and $b=\alpha _2$ in what follows. Define

$$\begin{aligned} G(a,b) = \sin ^3(b) f(a,b) - \sin ^3(a) f(b,a) \quad \quad A:=(a,b) \in E, \end{aligned}$$

where the region E is given by (2.5). Our goal is to establish that $G(A) > 0$ for all $A \in E$. Let X(A) be the vector field defined in (2.6) and note that $X(A)=\nabla G(A)$. Consider the planar, autonomous ODE defined by

$$\begin{aligned} \dot{A}(t)&= X(A(t)) \quad t \ge 0, \\ A(0)&= A, \end{aligned}$$

where A is fixed and belongs to E, and let the associated (forward) trajectory be ${\phi ^{+}_{\scriptscriptstyle {A}}}:=\{A(t): \ t \ge 0\}.$ According to part (iv) of Lemma 2.3, the system has no rest points in E, and, by parts (i)–(iii), the flow field X(A) is such that trajectories can enter E only by crossing OB or OC. See Fig. 2. Thus, there is a point $A'=A(t_0)$, say, with $t_0<0$, belonging to the backward trajectory ${\phi ^{-}_{\scriptscriptstyle {A}}}:=\{A(t): \ t \le 0\}$ and which lies either on OB or OC. Now define for $t \ge t_0$ the function $g(t):=G(A(t))$ and note that

$$\begin{aligned} \dot{g}(t)&= \nabla G(A(t)) \cdot X(A(t)) \nonumber \\&= |X(A(t))|^2. \end{aligned}$$

(2.7)

By part (iv) of Lemma 2.3, $\dot{g}(t) > 0$ for all $t > t_0$, and, by inspection, $g(t_0)=G(A')=0$. Hence $G(A)=g(0)>g(t_0)=0$. $\square$

In view of the above, it is tempting to suppose that, for a convex polygon $\Omega$, $\max _{\partial \Omega }\zeta$ occurs at the vertex with the most acute angle. However, this is not the case. For instance, if we consider a quadrilateral obtained from an isosceles triangle I whose repeated angle is $\tau >\frac{\pi }{3}$ then, by Proposition 2.1, we know that

$$\begin{aligned} \max _{x \in I} \zeta (x;I) = \zeta (V;I), \end{aligned}$$

where the vertex V of I is associated with the angle $\gamma <\frac{\pi }{3}$. By splitting V into two new vertices $V^{\pm }$, say, and forming the quadrilateral Q, as shown in Fig. 3, we can argue by continuity of the integral and symmetry that we must have

$$\begin{aligned} \max _{x \in Q}\zeta (x;Q) = \zeta (V^{\pm };Q) \end{aligned}$$

for sufficiently small $\epsilon$. By inspection, the angles associated with $V_{\pm }$ exceed $\tau$, so the maximum of $\zeta (\cdot ;Q)$ is not associated with the vertex (or vertices) with the ‘most acute’ angle. The discussion above constitutes a proof of the following result.

Fig. 2

The flow field associated with the ODE $\dot{A}=X(A)$. The region E is bounded by the red triangle (Color figure online)

Fig. 3

An illustration of the assertion that, in the case of a quadrilateral, the maximum of $\zeta$ need not occur at the vertex (or vertices) associated with the smallest angle

Proposition 2.5

There exist convex quadrilaterals Q with the property that $\max _{\partial Q}\zeta$ does not occur at the vertex (or vertices) with the smallest interior angle.

3 Determining $C(\Omega )=\max \zeta (\cdot ;\Omega )$ When $\Omega$ is a General Convex Polygon

When $\Omega$ is a triangle, $\max \zeta (\cdot ;\Omega )$ can be determined directly using Proposition 2.1, or by applying Propositions 2.2 and 2.4. By contrast, the situation with general convex polyhedra $\Omega$ is less straightforward, as, for example, the result of Proposition 2.5 presages. Indeed, even for quadrilaterals there seems to be no simple criterion which, in general, reliably determines the ordering of the $\zeta (V_i,\Omega )$, despite the fact that an explicit expression for $\zeta (V_i,\Omega )$ in terms of the position vectors of the vertices is always available. Our approach is therefore to (i) provide a geometric criterion capable of identifying $\max _i \zeta (V_i)$ for a large class of convex polygons, and, for general convex polygons, (ii) to identify a variety of inequalities that bound $\max _i \zeta (V_i)$ from above and below in terms of easily-calculated quantities. The motivation for (ii) is twofold: as well as being of practical use, the various inequalities are intrinsically geometrical in nature, and, as such, give insights into the structure of $\zeta (\cdot ,\Omega )$. For (i), see Subsects. 3.1 and 3.2 below; for (ii), see Subsects. 3.3 and 3.4. Finally, in Subsect. 3.5, we moot a third approach to ordering the values $z(V_i;\Omega )$ for a convex polygon $\Omega$. It turns out that what we term ‘magic points’ always exist in the case that $\Omega$ is a triangle, seem always to exist when $\Omega$ is a quadrilateral, and generally seem not to always exist for polygons with five vertices or more.

3.1 Calculating z for a Planar Polygon

Let Q be a planar polygon with vertices $V_i$ for $i=1,\ldots ,k$, labelled anticlockwise with increasing i, and let

$$\begin{aligned} z(x;Q)=\int _{Q} |x-y| \, d\mathcal {L}^2(y). \end{aligned}$$

Thus $z(x,Q)=\mathcal {L}^2(Q)\zeta (x,Q)$, and so, for fixed $\mathcal {L}^2(Q)$, it is sufficient to maximize z in order to maximize $\zeta$.

In order to calculate $z(V_i;Q)$ for a given i in ${1,\ldots ,k}$, we first triangulate Q according to

$$\begin{aligned} Q =\bigcup _{j, j+1\ne i} T_j \end{aligned}$$

where $T_j$ is the triangle $V_iV_jV_{j+1}$, and note the obvious relation

$$\begin{aligned} z(V_i,Q) =\sum _{j,j+1 \ne i} z(V_i,T_j). \end{aligned}$$

(3.1)

The following lemma provides an expression for $z(V_1,T)$ in terms of the position vectors of the vertices $V_1,V_2,V_3$ of a given triangle T. When used repeatedly in (3.1), the formula (3.5) for $z(V_1,Q)$ emerges. In the lemma, we make use of the natural embedding of $\mathbb {R}^2$ into $\mathbb {R}^3$ given by $\tilde{v}:=(v^1,v^2,0)$ for $v=(v_1,v_2) \in \mathbb {R}^2$. We then define for any $v,w \in \mathbb {R}^2$ the scalar quantity

$$\begin{aligned} v \, \textbf{x} \,w:= (\tilde{v} \times \tilde{w}) \cdot e_3, \end{aligned}$$

where $e_3=(0,0,1)$ and $\times$ denotes the classical cross product in $\mathbb {R}^3$.

Lemma 3.1

Let $T=V_1V_2V_3$ be a triangle whose vertices have been labelled anticlockwise with increasing index. Let $\alpha =\angle V_3V_1V_2$ and $\beta =\angle V_3V_2V_1$, and define the function $h:(0,\pi ) \rightarrow \mathbb {R}$ by

$$\begin{aligned} h(\theta )=\ln (\tan (\theta /2))-\frac{\cos \theta }{\sin ^2\theta } \end{aligned}$$

(3.2)

Then

$$\begin{aligned} z(V_1,T)=\frac{|V_2-V_1|^3\sin ^3\beta }{6}(h(\alpha +\beta )-h(\beta )). \end{aligned}$$

(3.3)

Proof

We regard $V_1$ as the origin of a set of plane polar coordinates $(R,\theta )$ in which $\theta =0$ corresponds to the direction $V_2':=V_2-V_1$. For brevity, we henceforth write $X'=X-V_1$ for any vector X, and then drop the primes for clarity. Now, points $X_0$ on the line $[V_2,V_3]$ obey

$$\begin{aligned} \widetilde{X_0} \, \textbf{x} \,\widetilde{W}&= \widetilde{V_2} \, \textbf{x} \,\widetilde{W}, \end{aligned}$$

(3.4)

where $W:=V_3-V_2$. Writing $X_0=R(\theta )(\cos \theta ,\sin \theta )$ and solving (3.4) for $R(\theta )$ yields

$$\begin{aligned} R(\theta ) =\frac{|\widetilde{V_2} \, \textbf{x} \,\widetilde{V_3}|}{|V_3-V_2|\sin (\theta +\beta )}, \end{aligned}$$

and hence the description

$$\begin{aligned} T=\{R (\cos \theta , \sin \theta ): \ 0 \le R \le R(\theta ), \ 0 \le \theta \le \alpha \}. \end{aligned}$$

Thus

$$\begin{aligned} z(V_1,T)&=\int _{T}|X|\, dx \\ {}&= \int _0^{\alpha _1} \frac{R^3(\theta )}{3} \, d\theta \\&= \frac{|\widetilde{V_2} \, \textbf{x} \,\widetilde{V_3}|^3}{6|V_3-V_2|^3}(h(\alpha +\beta )-h(\beta )). \end{aligned}$$

Finally, note that, on reverting to the original coordinates,

$$\begin{aligned} |\widetilde{V_2} \, \textbf{x} \,\widetilde{V_3}|=|V_2-V_1||V_3-V_2|\sin \beta , \end{aligned}$$

which leads directly to

$$\begin{aligned} z(V_1,T)&= \frac{|V_j-V_1|^3\sin ^3 \beta }{6}(h(\alpha +\beta )-h(\beta )). \end{aligned}$$

$\square$

Using this and (3.1), we find that for a convex polygon with k vertices,

$$\begin{aligned} z(V_1;Q)&= \sum _{j=1}^{k-2} \frac{|V_j-V_1|^3\sin ^3\beta _j}{6}(h(\alpha _j+\beta _j)-h(\beta _j)), \end{aligned}$$

(3.5)

where $\alpha _j:=\angle V_{j+2}V_1 V_{j+1}$ and $\beta _{j}:=\angle V_{j+2}V_{j+1}V_1$ for $j=1,\ldots ,k-2$. This expression is easily adapted to calculate $z(V_i;Q)$ for general i, and we omit the details for brevity.

3.2 SIVs and IRRs

The expression (3.5), while useful for calculations in specific cases, does not seem to make much easier the task of identifying the vertex (or vertices) that maximize z. There are circumstances, however, in which the original definition of $z(\cdot ;Q)$ and the specific geometry of Q combine and allow us to say more, effectively by-passing the use of (3.5). We shall need the following two definitions, which are given in terms of a general (convex) polyhedron $\Omega \subset \mathbb {R}^d$ with vertices $V_1,\ldots ,V_k$. We suppose that $k\ge d+1$ in order that $\Omega$ has positive measure in $\mathbb {R}^d$.

Definition 3.2

(SIV) For $i\ne j$, $1 \le i, j \le k$, let $PB(V_i,V_j)$ represent the hyperplane that bisects the part-line $[V_i,V_j]$ and is orthogonal to the vector

$$\begin{aligned} n_{ij}:= (V_j-V_i)/|V_i-V_j|, \end{aligned}$$

that is

$$\begin{aligned} PB(V_i,V_j):=\frac{V_i+V_j}{2}+n_{ij}^{\perp }. \end{aligned}$$

(3.6)

Then $PB(V_i,V_j)$ is the intersection of the two half-spaces

$$\begin{aligned} H_{ij}^{\pm }:=\left\{ X: \pm \left( X-\frac{V_i+V_j}{2}\right) \cdot n_{ij} \ge 0\right\} , \end{aligned}$$

(3.7)

and we say that $V_i$ is a singly isolated vertex (SIV) relative to $V_j$ if all vertices of $\Omega$ other than $V_i$ lie in $H_{ij}^+$ (Fig. 4).

Fig. 4

Left: $V_1$ is a Singly Isolated Vertex (SIV) relative to $V_2$. Right: $V_1$ creates an Internally Reflected Region (IRR) relative to $V_2$

Implicit in the definition of $H_{ij}^{\pm }$ is the inclusion $V_i \in H_{ij}^{-}$, so, put simply, if $V_i$ is an SIV relative to $V_j$ then $V_i$ lies on one side of $PB(V_i,V_j)$ and the rest of the vertices of $\Omega$ lie on the other side. With this in mind, the following characterization of SIVs is immediate.

Proposition 3.3

Let $\Omega \subset \mathbb {R}^d$ be a convex polyhedron with vertices $V_1,\ldots ,V_k$. Then $V_i$ is an SIV relative to $V_j$ iff $|V_r - V_i| \ge |V_r-V_j|$ for all $r=1,\ldots ,k$, $r\ne i$.

Closely related to SIVs are internally reflected regions, or IRRs for short, whose definition is as follows.

Definition 3.4

(IRR) For $i\ne j$, $1 \le i, j \le k$, and with the half-spaces $H_{ij}^{\pm }$ as described in (3.7), form the two sets

$$\begin{aligned} \Omega _{ij}^{\pm }:= \Omega \cap H_{ij}^{\pm }. \end{aligned}$$

(3.8)

Let $\rho _{ij}$ be the affine map defined by

$$\begin{aligned} \rho _{ij}(y)= (\textbf{1}- 2 n_{ij} \otimes n_{ij})(y-m_{ij}) + m_{ij} \quad \quad y\in \mathbb {R}^d, \end{aligned}$$

where $m_{ij}:=(V_i+V_j)/2$. Then $V_i$ creates an internally reflected region (IRR) relative to $V_j$ if

$$\begin{aligned} \rho _{ij}(\Omega _{ij}^{-})\subset \Omega _{ij}^+. \end{aligned}$$

(3.9)

Notice that $\rho _{ij}(y)$ is a reflection of y in $PB(V_i,V_j)$, and so $V_i$ creates an IRR relative to $V_j$ when the reflection of $\Omega _{ij}^{-}$ in $PB(V_i,V_j)$ is a subset of $\Omega _{ij}^+$. It is not difficult to see that SIVs are particular instances of IRRs, as follows.

Proposition 3.5

Let $\Omega \subset \mathbb {R}^d$ be a convex polyhedron with vertices $V_1,\ldots ,V_k$, and suppose $V_i$ is an SIV relative to $V_j$. Then $V_i$ creates an IRR relative to $V_j$.

Proof

If $V_i$ is an SIV relative to $V_j$ then, by definition, $V_i$ is the only vertex in $H_{ij}^{-}$. Let S be the set formed by intersecting the closed faces of $\Omega$ containing $V_i$ with $PB(V_i,V_j)$. In particular, S is a closed, convex subset of $\Omega$, and, since all the other vertices of $\Omega$ lie in $\Omega _{ij}^+$, we must have

$$\begin{aligned} (S \cup V_i)^c = \Omega _{ij}^{-}. \end{aligned}$$

The reflection $\rho _{ij}$ defined above maps $V_i$ to $V_j$ and fixes every point of S. By a direct calculation,

$$\begin{aligned} \rho _{ij}( (S \cup V_i)^c) = (S \cup V_j)^c \subset H_{ij}^+, \end{aligned}$$

and the inclusions $S\subset \Omega$, $V_j \subset \Omega$ imply $(S \cup V_j)^c \subset \Omega$ by the convexity of $\Omega$. Hence $\rho _{ij}(\Omega _{ij}^{-})\subset \Omega _{ij}^{+}$, as required. $\square$

Our main application for both SIVs and IRRs is the following, which, in view of Proposition 3.5, we need only prove in the case of an IRR in order to conclude that it applies to any SIV too.

Lemma 3.6

Let $\Omega \subset \mathbb {R}^d$ be a convex polyhedron with vertices $V_1,\ldots ,V_k$, and suppose that $V_i$ creates an IRR relative to $V_j$. Then $z(V_i;\Omega )\ge z(V_j;\Omega )$.

Proof

Dropping the second argument of z and writing $dy=d\mathcal {L}^2(y)$, $dY = d\mathcal {L}^2(Y)$ for brevity, start by noting

$$\begin{aligned} z(V_i) = \int _{\Omega } |V_i-y| \, dy&= \int _{\Omega _{ij}^{-}} |V_i-y| \, dy \nonumber \\&\quad +\int _{\rho _{ij}(\Omega _{ij}^{-})}|V_i-y| \, dy \nonumber \\&\quad +\int _{\Omega _{ij}^+\setminus \rho _{ij}(\Omega _{ij}^{-})}|V_i-y| \, dy, \end{aligned}$$

(3.10)

where $\Omega _{ij}^{\pm }$ as defined in (3.8) and where the inclusion (3.9) has been used. Similarly,

$$\begin{aligned} z(V_j)&= \int _{\Omega _{ij}^{-}} |V_j-Y| \, dY + \int _{\rho _{ij}(\Omega _{ij}^{-})}|V_j-Y| \, dY \nonumber \\&\quad +\int _{\Omega _{ij}^+\setminus \rho _{ij}(\Omega _{ij}^{-})}|V_j-Y| \, dY. \end{aligned}$$

(3.11)

In the first integral of (3.11), let $Y=\rho _{ij}(y)$ and note that $\Omega _{ij}^{-}=\rho _{ij}(\rho _{ij}(\Omega _{ij}^{-}))$ and $\rho _{ij}(V_i)=V_j$. This gives

$$\begin{aligned} \int _{\Omega _{ij}^{-}} |V_j-Y| \, dY = \int _{\rho _{ij}(\Omega _{ij}^{-})}|V_i-y| \, dy,\end{aligned}$$

and the same change of variables produces

$$\begin{aligned} \int _{\rho _{ij}(\Omega _{ij}^{-})} |V_j-Y| \, dY = \int _{\Omega _{ij}^{-}}|V_i-y| \, dy \end{aligned}$$

in the second integral of (3.11). Hence,

$$\begin{aligned} z(V_j)&=\int _{\Omega _{ij}^{-}}|V_i-y| \, dy + \int _{\rho _{ij}(\Omega _{ij}^{-})}|V_i-y| \, dy \nonumber \\&\quad +\int _{\Omega _{ij}^+\setminus \rho _{ij}(\Omega _{ij}^{-})}|V_j-Y| \, dY,\end{aligned}$$

(3.12)

the first two terms of which correspond exactly to the first two terms of (3.10). It therefore follows from (3.10) to (3.12) that

$$\begin{aligned} z(V_i) -z(V_j)&= \int _{\Omega _{ij}^+\setminus \rho _{ij}(\Omega _{ij}^{-})}|V_i-y| -|V_j-y| \, dy. \end{aligned}$$

(3.13)

To conclude, note that $|V_i-y| \ge |V_j-y|$ for all y in the half-space $H_{ij}^+$, and so, in particular, the integrand of the right-hand side of (3.13) is nonnegative in the region of integration.$\square$

As a corollary, we find that the following condition is capable of identifying a (or the) maximizing vertex.

Corollary 3.7

Let $\Omega \subset \mathbb {R}^d$ be a convex polyhedron with vertices $V_1,\ldots ,V_k$, and suppose that for each $j \in \{2,\ldots ,k\}$ $V_1$ creates an IRR relative to $V_j$. Then $z(V_1)=\max _{\Omega }z$.

Proof

By Lemma 3.6, we have $z(V_1)\ge z(V_j)$ for $j=2,\ldots k$, and since $\max _{\Omega }z$ is attained at a vertex, the claim follows. $\square$

Corollary 3.8

Let T be a triangle with (internal) angle $\alpha _i$ at vertex $V_i$ for $i=1,2,3$, and suppose that $\alpha _1 \le \alpha _2 \le \alpha _3$. Then $z(V_1,T) \ge z(V_2,T) \ge z(V_3,T)$, and, in particular, we recover the result of Proposition 2.1.

Proof

By inspection, if $\alpha _1 \le \alpha _2$ then $V_1$ is an SIV relative to $V_2$, and, similarly, $V_2$ is an SIV relative to $V_3$. Since, by Proposition 3.5, $V_1$ then creates an IRR relative to $V_2$, as does $V_2$ relative to $V_3$, Lemma 3.6 implies that $z(V_1,T) \ge z(V_2,T) \ge z(V_3,T)$. $\square$

Remark 3.9

Concrete instances of quadrilaterals to which Corollary 3.7 applies can be constructed as follows. First, Proposition 3.3 tells us that $V_1$ is an SIV relative to each of $V_2,V_3$ and $V_4$ if and only if

$$\begin{aligned} |V_3-V_1|&\ge \max \{|V_3-V_2|,|V_3-V_4|\}, \\ |V_4-V_1|&\ge \max \{|V_4-V_2|,|V_4-V_3|\}, \\ |V_2-V_1|&\ge \max \{|V_2-V_3|,|V_2-V_4|\}. \end{aligned}$$

By a direct calculation, all three inequalities can be satisfied by any cyclic quadrilateral with vertices $V_1=(-1,0)$, $V_2=(\cos \alpha , -\sin \alpha )$, $V_3=(1,0)$ and $V_4=(\cos \alpha , \sin \alpha )$ as long as $0 \le \alpha \le \pi /3$. In fact, a numerical computation shows that $z(V_1)=\max _{\Omega }z$ remains true for $0 \le \alpha < \pi /2$. $\diamond$

3.3 Upper and Lower Bounds on $\max \zeta (\cdot ;\Omega )$ When $\Omega$ is a Polyhedron

The following result, which applies to general convex 2n-gons¹$\Omega$, provides us with a first, simple ‘geometric’ lower bound on $\max _{\Omega }z$. The lower bound is geometric in the sense that it depends on the codiameter of $\Omega$, $\textrm{codiam}\,(\Omega )$, defined by

$$\begin{aligned} \textrm{codiam}\,(\Omega ):=\sup _{P \in P(2n)} \sum _{(i,j) \in P} |V_i-V_j|, \end{aligned}$$

(3.14)

where P(2n) represents the class of all possible partitions P, say, of the set $\{1,\ldots ,2n\}$ into n distinct pairs (i, j) with $i \ne j$, and $V_1,\ldots ,V_{2n}$ are the vertices of $\Omega$. An alternative definition of $\textrm{codiam}\,(\Omega )$ can be given in terms of the set $S\subset S_{2n}$ of permutations on 2n elements, where

$$\begin{aligned} S:=\{\sigma \in S_{2n}: \ \sigma =(i_1i_2)(i_3i_4)\ldots (i_{2n-1}i_{n}), \ \{i_j\}_{j=1\ldots 2n} \simeq \{1,\ldots ,2n\}\}. \end{aligned}$$

(3.15)

Here, $E \simeq F$ is used to indicate that the sets E and F can be put into bijection. In terms of permutations,

$$\begin{aligned} \textrm{codiam}\,(\Omega ):=\frac{1}{2}\sup _{\sigma \in S} \sum _{i=1}^{2n} |V_i-V_{\sigma (i)} |. \end{aligned}$$

(3.16)

For a concrete instance of $\textrm{codiam}\,(Q)$ when Q is a planar quadrilateral, see (3.18) below. In the following proof, we use the abbreviation $z_i:=z(V_i)$ for all i.

Proposition 3.10

Let $\Omega \subset \mathbb {R}^d$ be a bounded, convex 2n-gon in $\mathbb {R}^d$. Then

$$\begin{aligned} \max _{\Omega } \zeta \ge \frac{\textrm{codiam}\,(\Omega )}{2n}. \end{aligned}$$

(3.17)

Proof

Let $\sigma$ be a permutation in the set S, as defined in (3.15), and let i belong to $\{1,\ldots ,n\}$. Since for any Y in $\Omega$ it holds that $|V_i-Y|+|V_{\sigma (i)}-Y| \ge |V_i-V_{\sigma (i)}|$, integrating this expression with respect to such Y yields $z_i+z_{\sigma (i)} \ge \mathcal {L}^d(\Omega ) |V_i-V_{\sigma (i)}|$. Summing over $i=1,\dots ,2n$ gives

$$\begin{aligned} \frac{2}{\mathcal {L}^d(\Omega )} \sum _{j=1}^{2n} z_j \ge \sum _{i=1}^{2n} |V_i-V_{\sigma (i)}|. \end{aligned}$$

The left-hand side is independent of $\sigma$ and is bounded above by $4n \max _{\Omega } \zeta$, so by taking the supremum of the right-hand side over all $\sigma$ in S, applying (3.16) and then rearranging, we obtain (3.17). $\square$

In the case that $\Omega$ is a quadrilateral $Q \subset \mathbb {R}^2$ whose vertices are labelled anticlockwise $V_1,\ldots ,V_4$, $\textrm{codiam}\,(Q)$ becomes

$$\begin{aligned} \textrm{codiam}\,(Q)&=\max \{|V_1-V_2|+|V_3-V_4|,|V_1-V_4| \nonumber \\&\quad +|V_2-V_3|,|V_1-V_3|+|V_2-V_4|\}. \end{aligned}$$

(3.18)

But since it is always the case that the sum of the lengths of the diagonals of a convex quadrilateral exceeds the sum of the lengths of any pair of opposite sides, that is

$$\begin{aligned} |V_1-V_3|+|V_2-V_4|&\ge |V_1-V_2|+|V_3-V_4| \\ |V_1-V_3|+|V_2-V_4|&\ge |V_1-V_4|+|V_2-V_3|, \end{aligned}$$

it must be that $\textrm{codiam}\,(Q)=|V_1-V_3|+|V_2-V_4|$. This leads to the following straightforward lower bound on $\max _{Q} z$.

Proposition 3.11

Let Q be a planar quadrilateral, and let $\theta$ be the acute angle formed at the point where the diagonals $V_1V_3$ and $V_2V_4$ cross. Then

$$\begin{aligned} \max _{Q} \zeta&\ge \sqrt{\frac{\mathcal {L}^2(Q) \, \textrm{cosec}(\theta )}{2}}. \end{aligned}$$

(3.19)

Proof

By Proposition 3.10 and the expression for $\textrm{codiam}\,(Q)$ derived above, it is in particular the case that

$$\begin{aligned} \max _Q \zeta&\ge \frac{1}{4}\left( |V_1-V_3|+|V_2-V_4|\right) \\&= \frac{1}{4}\left( |V_1-V_3|+\frac{2\,\textrm{cosec}(\theta )\mathcal {L}^2(Q)}{|V_1-V_3|}\right) \\&\ge \sqrt{\frac{\mathcal {L}^2(Q) \, \textrm{cosec}(\theta )}{2}}. \end{aligned}$$

In the above, we have used the well-known formula $\mathcal {L}^2(Q) =\frac{|V_2-V_4||V_1-V_3|\sin \theta }{2}$. $\square$

Remark 3.12

We remark that some information is lost in both lower bounds (3.17) and (3.19). To see this, let S be the unit square in $\mathbb {R}^2$ and note that by direct calculation we have

$$\begin{aligned} \max _S z&= \frac{\sqrt{2}}{3} + \frac{\ln (1+\sqrt{2})}{3} \sim 0.7652, \\ \frac{\textrm{codiam}\,(S)}{4}&= \frac{1}{\sqrt{2}} \sim 0.7071, \\ \sqrt{\frac{\textrm{cosec}(\theta _S)}{2}}&= \frac{1}{\sqrt{2}}. \end{aligned}$$

This is natural when one examines the case of equality in the proof of Proposition 3.10. $\diamond$

According to the calculation above, for a convex and planar quadrilateral Q it holds that $\textrm{codiam}\,(Q)=d_{13}+d_{24}$, where $d_{ij}:=|V_i-V_j|$ for all i, j. Numerical experiments suggest that the following bounds should then hold true for general quadrilaterals Q of unit area

$$\begin{aligned} N_0(d_{13}+d_{24}) \le \max _Q z \le N_1(d_{13}+d_{24}), \end{aligned}$$

(3.20)

where $N_1$ and $N_0$ are close to $\frac{2}{3}$ and $\frac{1}{4}$ respectively. This can be established analytically as follows. Firstly, by applying [1, Proposition 3.1] to $Q \subset \mathbb {R}^2$, which states that

$$\begin{aligned} \frac{1}{2}\textrm{diam}\,(Q) \le \max _{Q}z \le \frac{2}{3} \textrm{diam}\,(Q), \end{aligned}$$

(3.21)

together with the obvious inequalities $\textrm{diam}\, (Q) \le \textrm{codiam}\,(Q) \le 2 \,\mathrm {diam(}Q)$, we see that

$$\begin{aligned} \frac{1}{4} \textrm{codiam}\,(Q) \le \max _{Q}z \le \frac{2}{3} \textrm{codiam}\,(Q) \end{aligned}$$

(3.22)

should hold. To see that this is sharp we argue as follows, starting with the prefactor $N_1$ in (3.23). A calculation shows that when Q is a quadrilateral such that $d_{13}=2/\epsilon$, $d_{24}=\epsilon<<1$ and in which the diagonals are orthogonal and cross at a point very near the vertex $V_3$, say, we have $z(V_1) \simeq \frac{4}{3\epsilon }$. Since in this case $\textrm{codiam}\,(Q)=\epsilon +\frac{2}{\epsilon }$, (3.20) requires that

$$\begin{aligned} \frac{4}{3\epsilon } \lesssim N_1 \left( \epsilon +\frac{2}{\epsilon }\right) \end{aligned}$$

(3.23)

for all sufficiently small $\epsilon >0$. Rearranging and then letting $\epsilon \rightarrow 0$ shows that $N_1 \ge \frac{2}{3}$, which in conjunction with (3.22) proves that the ‘best’, i.e. smallest, $N_1$ must be the claimed $\frac{2}{3}$. As for the lower bound, note that the sets $Q_{\epsilon }:=[0,\epsilon ^{-1}]\times [0,\epsilon ]$ are such that $\mathcal {L}^2(Q_{\epsilon })=1$,

$$\begin{aligned} \max _{Q_{\epsilon }}z&\le \frac{1}{2\epsilon } + O(\epsilon ) \ \text {and} \\ \textrm{codiam}\,(Q_{\epsilon })&= \frac{2}{\epsilon } + O(\epsilon ), \end{aligned}$$

so that

$$\begin{aligned} N_0 \le \frac{\max _{Q_{\epsilon }}z}{\textrm{codiam}\,(Q_{\epsilon })} \le \frac{1}{4}+O(\epsilon ^2). \end{aligned}$$

Finally, since $\textrm{codiam}\,(Q) \ge \textrm{diam}\,(Q)$, the tightest possible bounds on $\max _Q z$ arising from (3.21) to (3.22) are as follows.

Proposition 3.13

Let Q be a bounded, convex and planar quadrilateral. Then

$$\begin{aligned} \frac{1}{4} \textrm{codiam}\,(Q) \le \max _{Q}z \le \frac{2}{3} \textrm{diam}\,(Q). \end{aligned}$$

In the case of polygons with odd numbers of vertices, a slightly different approach is needed in order to generate ‘geometric’ lower bounds on $\max _{\Omega }z$. We illustrate the process first with the simplest possible shape: a triangle T, say, in the plane. Letting the vertices of T be $V_1,V_2, V_3$, labelled anticlockwise as usual, we define the quantities

$$\begin{aligned} A(1,2)&:=\left| V_3-\frac{V_1+V_2}{2}\right| + \frac{\pi }{3\mathcal {L}^2(T)} r^3\left( \frac{V_1+V_2}{2}\right) , \end{aligned}$$

(3.24)

$$\begin{aligned} A(1,3)&:= \left| V_2-\frac{V_1+V_3}{2}\right| + \frac{\pi }{3\mathcal {L}^2(T)} r^3\left( \frac{V_1+V_3}{2}\right) , \end{aligned}$$

(3.25)

$$\begin{aligned} A(2,3)&:= \left| V_1-\frac{V_2+V_3}{2}\right| + \frac{\pi }{3\mathcal {L}^2(T)}r^3\left( \frac{V_2+V_3}{2}\right) , \end{aligned}$$

(3.26)

where the function $r: \partial T \setminus \{V_1,V_2,V_3\} \rightarrow \mathbb {R}$ is given by

$$\begin{aligned} r(y):=\sup \{t>0: \textrm{SC}(y,t) \subset T\} \quad y \in \partial T\setminus \{V_1,V_2,V_3\}. \end{aligned}$$

(3.27)

The notation $\textrm{SC}(y,t)$ refers to the semicircle of radius t and centre y, and, as such, we can think of r(y) as a type of ‘maximum inscribed radius’ relative to the (non-vertex) boundary point y. We make use of these definitions as follows.

Proposition 3.14

Let $T\subset \mathbb {R}^2$ be a triangle with vertices $V_1,V_2,V_3$, and let the quantities A(i, j) for $1\le i, j \le 3$, $i \ne j$, be given by (3.24)–(3.26). Then

$$\begin{aligned} \max _T \zeta&\ge \frac{\max \{A(1,2),A(1,3),A(2,3)\}}{3}. \end{aligned}$$

(3.28)

Proof

First note that, according to the definition of r in (3.27), it holds for any non-vertex $y \in \partial T$ that

$$\begin{aligned} z(y)&= \int _T |y-Y| \,dY \\&\ge \int _{SC(y,r(y))} |y-Y| \, dY \\&= \frac{\pi }{3}r^3(y). \end{aligned}$$

The last line is obtained by integrating in polar coordinates, the details of which we omit.

Now let $Y \in T$ and consider

$$\begin{aligned} |Y-V_1|+|V_2-Y|+|V_3-Y|&\ge |Y-V_1|+|V_2+V_3-2Y| \end{aligned}$$

which, on integration over Y in T, yields,

$$\begin{aligned} z_1+z_2+z_3&\ge \int _T |Y-V_1| + \left| \frac{V_2+V_3}{2}-Y\right| \, dY + z\left( \frac{V_2+V_3}{2}\right) \\&\ge \int _T \left| \frac{V_2+V_3}{2}-V_1\right| \, dY + \frac{\pi }{3} r^3\left( \frac{V_2+V_3}{2}\right) = \mathcal {L}^2(T) A(2,3). \end{aligned}$$

Arguing similarly in the other cases, we see that

$$\begin{aligned} \frac{1}{\mathcal {L}^2(T)} (z_1+z_2+z_3) \ge \max \{A(1,2),A(1,3),A(2,3)\}, \end{aligned}$$

which, since the left-hand side of this last inequality is bounded above by $3\max _{T}\zeta$, immediately implies (3.28). $\square$

Remark 3.15

Numerical experiments show that

$$\begin{aligned} \frac{2}{3} \max \{A(1,2),A(1,3),A(2,3)\}&\ge \max _{T}z \nonumber \\&\ge N \max \{A(1,2),A(1,3),A(2,3)\} \end{aligned}$$

(3.29)

ought to hold in general, where

$$\begin{aligned} N=\frac{12 \ln (3)+16}{3\pi +24\sqrt{3}}. \end{aligned}$$

The constant $N\simeq 0.572$ appearing as a prefactor in the lower bound obeys $N=z(V_1;E)/A(2,3)$, where E is an equilateral triangle of unit area. It is clearly an improvement on (3.28), and it is natural to conjecture that the quantity

$$\begin{aligned} \frac{\max _{T}z}{\max \{A(1,2),A(1,3),A(2,3)\}} \end{aligned}$$

is decreasing with increasing symmetry of T, perhaps via a Steiner symmetrization argument within the class of triangles. $\diamond$

In the case of isosceles triangles at least, the direct calculation given below shows that the maximizer of $\max \{A(1,2),A(1,3),A(2,3)\}$ also predicts which of $z_1$ and $z_2(=z_3)$ is maximal. Bearing Proposition 2.1 in mind, it is enough to show that the maximizer is capable of ordering the angles $\alpha _1=\angle V_3V_1V_2$ and $\alpha _2=\angle V_1V_2V_3$.

Let T be an isosceles triangle with $\delta := |V_1V_2|=|V_1V_3|$, and let the angles $\alpha _1$ and $\alpha _2$ be as described above. By symmetry, $A(1,3)=A(1,2)$, so the lower bound in (3.28) becomes $A(1,2) \vee A(2,3)$, where $p \vee q$ is shorthand notation for $\max \{p,q\}$. Our assertion is that $A(1,2) < A(2,3)$ if and only if $\alpha _1<\alpha _2$. To see it, first let $\theta =\frac{\alpha _1}{2}$ and note that

$$\begin{aligned} r\left( \frac{V_2+V_3}{2}\right)&= \delta \sin \alpha _2 \sin \theta \\ r\left( \frac{V_1+V_2}{2}\right)&= \frac{\delta }{2}\sin \alpha _1. \end{aligned}$$

Since $|T|=1$, it must be that $\delta ^2\sin \theta \cos \theta =1$, which together with the fact that $\alpha _2=\pi /2-\theta$, gives

$$\begin{aligned} r\left( \frac{V_2+V_3}{2}\right)&= \delta \cos \theta \sin \theta = \frac{1}{\delta }\\ r\left( \frac{V_1+V_2}{2}\right)&= \frac{\delta }{2}\sin \alpha _1 = \frac{1}{\delta }. \end{aligned}$$

Referring to (3.24) and (3.26), it is evident that the ‘inscribed radius’ terms are identical, and therefore that $A(1,2) < A(2,3)$ if and only if

$$\begin{aligned} \left| V_3-\frac{V_1+V_2}{2}\right|&< \left| V_1-\frac{V_2+V_3}{2}\right| \end{aligned}$$

In terms of $\theta$, the latter reads

$$\begin{aligned} \frac{\delta }{2}\sqrt{1+8 \sin ^2\theta } < \delta \cos \theta , \end{aligned}$$

which rearranges to $\sin \theta <\frac{1}{2}$, or simply $\alpha _1 < \pi /3$. Since this condition holds if and only if $\alpha _1 < \alpha _2$, it follows that

$$\begin{aligned} A(2,3)=A(1,2)\vee A(2,3) \ \text {if and only if} \ \alpha _1 < \alpha _2. \end{aligned}$$

By Proposition 2.1, and by recalling that $A(1,2)=A(1,3)$, we therefore have

$$\begin{aligned} A(2,3)=A(1,3)\vee A(2,3) \ \text {if and only if} \ z_1 > z_2. \end{aligned}$$

It is therefore the ‘omitted index’ in the arguments of the maximizer that predicts which vertex maximizes z. This leads us to conjecture that

Conjecture 3.16

For a general triangle,

$$\begin{aligned} A(i,j)=\max \{A(1,2), A(1,3), A(2,3)\} \ \text {if and only if} \ z_k \ge \max \{z_i, z_j\} \end{aligned}$$

where $\{i,j,k\}\simeq \{1,2,3\}$.

No exceptions to Conjecture 3.16 were found in the course of our numerical investigations.

3.4 The Role of the Centroid

Let Q be a convex, planar quadrilateral with vertices $V_1,\ldots ,V_4$, and define the centroid $\gamma (Q)$ of Q by

(3.30)

Also define for the same quadrilateral the quantity

$$\begin{aligned} D(Q):= \max _i \{|V_i-\gamma (Q)|\}. \end{aligned}$$

(3.31)

It turns out that D is a good indicator of $\max _i \zeta (V_i;\Omega )$ in the following sense.

Proposition 3.17

Let Q be a convex, planar quadrilateral with vertices $V_1,\ldots ,V_4$, and let $\gamma (Q)$ and D(Q) be as in (3.30) and (3.31) respectively. Then there is a constant $r_u$ such that

$$\begin{aligned} D(Q) \le \max _i \zeta (V_i;\Omega ) \le r_u D(Q). \end{aligned}$$

(3.32)

A numerical calculation gives $r_u \simeq 1.0822$ and a further numerical investigation suggests $r_u =(2-\sqrt{2}\ln (\sqrt{2}-1))/3$.

Proof

We prove the lower bound first. Let $V_i$ be any vertex of Q and write $\gamma$ for the centroid. Then, since

$$\begin{aligned} |V_i-y| \ge |V_i-\gamma |+\frac{V_i-\gamma }{|V_i-\gamma |} \cdot ({\gamma }-y) \quad y \in Q, \end{aligned}$$

an integration over Q leads immediately to $z(V_i;Q) \ge \mathcal {L}^2(Q)|V_i-\gamma |$. (Note that the second term vanishes by definition of the centroid.) Hence $\zeta (V_i;Q) \ge |V_i-\gamma |$ for $i=1,\ldots ,4$, from which we conclude that $Z(Q):=\max _i \zeta (V_i;\Omega )$ obeys $Z(Q) \ge |V_i-\gamma |$ for each i, and so, in particular, $Z(Q) \ge D(Q)$.

The upper bound results from a direct calculation, as follows. First, we parametrize Q using four angles $a_1,b_1,a_2,b_2$ and the length $\delta$ of one diagonal, as shown in Fig. 5. We choose coordinates so that $V_1=(0,0)$ and $V_3=(\delta ,0)$. Let

$$\begin{aligned} \rho (a,b)=\frac{\sin (a) \sin (b)}{\sin (a+b)} \quad 0< a+b< \pi \end{aligned}$$

and note that, in terms of $\rho$, we have

$$\begin{aligned} \mathcal {L}^2(Q) = \frac{\delta ^2}{2}(\rho (a_1,b_1)+\rho (a_2,b_2)). \end{aligned}$$

Using (3.5), we see that

$$\begin{aligned} z(V_1;Q)=\frac{\delta ^3}{6}(\sin ^3(b_1)f(a_1,b_1)+\sin ^3(b_2)f(a_2,b_2)), \end{aligned}$$

where $f(a,b)=h(a+b)-h(b)$ and h is given by (3.2). Hence

$$\begin{aligned} \zeta (V_1;Q)=\frac{\delta }{3}\left( \frac{\sin ^3(b_1)f(a_1,b_1)+\sin ^3(b_2)f(a_2,b_2)}{\rho (a_1,b_1)+\rho (a_2,b_2)}\right) . \end{aligned}$$

(3.33)

The constraints $0< a_1+a_2<\pi$ and $0<b_1+ b_2<\pi$ ensure that Q is a convex quadrilateral, and we apply these in what follows.

To calculate D(Q), henceforth D for brevity, we use the following expression for the centroid,

$$\begin{aligned} \gamma =\frac{\lambda _2 (V_2+V_3) +\lambda _1 (V_3+V_4)}{3}, \end{aligned}$$

(3.34)

which uses the notation

$$\begin{aligned} \lambda _i&:=\frac{\rho (a_i,b_i)}{\rho (a_1,b_1)+\rho (a_2,b_2)}, \quad i = 1,2. \end{aligned}$$

Next, letting

$$\begin{aligned} \sigma _i&= \frac{\sin (b_i)\cos (a_i)}{\sin (a_i+b_i)}, \quad i=1,2, \end{aligned}$$

and $\rho _i:=\rho (a_1,b_1)$ for $i=1,2$, we find that

$$\begin{aligned} \gamma = \frac{\delta }{3}(1+\lambda _2 \sigma _2 + \lambda _1 \sigma _1, \rho _1 -\rho _2), \end{aligned}$$

as well as $V_2 = \delta (\sigma _2, -\rho _2)$ and $V_4 = \delta (\sigma _1, \rho _1)$. Hence

$$\begin{aligned}&D=\frac{\delta }{3}\max \left\{ |(1+\lambda _2 \sigma _2 + \lambda _1 \sigma _1, \rho _1 -\rho _2)|,|(1+\lambda _1(\sigma _1-\sigma _2)-2\sigma _2,\rho _1+2\rho _2)|,\right. \end{aligned}$$

(3.35)

$$\begin{aligned}&\qquad \left. |(1+\lambda _2 (\sigma _2-\sigma _1)-2\sigma _1,-\rho _2-2\rho _1)|,|\lambda _1\sigma _1+\lambda _2\sigma _2-2,\rho _1-\rho _2) |\right\} \end{aligned}$$

(3.36)

$$\begin{aligned}&\quad =: \frac{\delta }{3}\overline{D}(a_1,b_1,a_2,b_2) \end{aligned}$$

(3.37)

Note that (i) $\overline{D}(Q)$, being the maximum of continuous functions, is itself continuous, and (ii) D(Q) is bounded away from 0. To see (ii), observe first that since each $V_i$ is an extreme point of Q and Q is polygonal,², there is a unit vector $n_i$, say, such that $(Y-V_i) \cdot n_i >0$ for a.e. Y in Q. This inequality, when integrated over Q, gives $(\gamma (Q)-V_i) \cdot n_i >0$, and hence $\gamma (Q) \ne V_i$. Assertion (ii) follows by using the definition of D(Q).

Now form the function $B = \frac{\zeta (V_1;Q)}{\overline{D}}$, which, in view of (3.33) and (3.35), can be written as

$$\begin{aligned} B(a_1,b_1,a_2,b_2)= \frac{\sin ^3(b_1)f(a_1,b_1)+\sin ^3(b_2)f(a_2,b_2)}{(\rho _1+\rho _2)\overline{D}(a_1,b_1,a_2,b_2)}. \end{aligned}$$

The task is to maximize $B(a_1,b_1,a_2,b_2)$ over positive $a_1,b_1,a_2,b_2$ such that

$$\begin{aligned} a_1 + b_1< \pi , \quad a_2+b_2< \pi , \quad a_1 + a_2< \pi , \quad \textrm{and} \ b_1 + b_2 < \pi . \end{aligned}$$

Denote by R, say, the set of such points $(a_1,b_1,a_2,b_2)$, and note that since B is continuous,

$$\begin{aligned} r_u:= \sup \{B(a_1,b_1,a_2,b_2): (a_1,b_1,a_2,b_2) \in \bar{R}\} \end{aligned}$$

exists (Fig. 6). Finally, we comment on the numerical value of $r_u$. In the numerical computation, $10^6$ uniformly-distributed [14, 15] random, quadrilaterals Q were generated and $\max _i{\zeta _i}/D(Q)$ was computed for each. The maximum value of this ratio found was 1.0822 and is an estimate for $r_u$. The quadrilateral corresponding to this value was drawn, and turned out to be close to a square. This suggests that $r_u =(2-\sqrt{2}\ln (\sqrt{2}-1))/3$, which is the exact value obtained for a square. $\square$

Fig. 5

The parametrization of Q used in the proof of Proposition 3.17

Fig. 6

A plot of $B(\frac{\pi }{4}, \frac{\pi }{4} + x, \frac{\pi }{4}, \frac{\pi }{4} + y)$ illustrating the apparent (at least local) maximum at the point $(a_1,b_1,a_2,b_2)=\left( \frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{4}\right) =:P$ when just two of the coordinates, in this case $a_2$ and $b_2$, are varied. Notice that B appears to be not differentiable at P

Remark 3.18

The fact that a maximizer of $B(Q):=\frac{Z(Q)}{D(Q)}$ appears to correspond to taking Q to be a square suggests strongly that an analytical proof of the upper bound in Proposition 3.17 ought to be possible via e.g. Steiner symmetrization. However, the effect of Steiner symmetrization on each of Z(Q) and D(Q) separately is to decrease them, so that B(Q) is a product of a decreasing function, Z(Q), and an increasing function, $\frac{1}{D(Q)}$, under symmetrization. If the same process is to increase B(Q), then $\frac{1}{D(Q)}$ must effectively ‘outcompete’ Z(Q).

Remark 3.19

We remark that the ‘overshoot’ of about $8\%$ implicit in estimate (3.32) demonstrates the limitations of D(Q) as a proxy for $\max _{Q}z$. The overshoot is smaller for triangles, and is progressively larger than $8\%$ for polygons with more sides.

The following table gives results for polygons $P_k$ with k sides, $k = 3, 4, 5, 6, 7$. In the table, $r_u(\textrm{num})$ was obtained by generating $10^6$ random, convex polygons $P_k$ of unit area, using the algorithm described in [14], and recording the maximum value of $r_u(\textrm{num}) = B(P_k)$. For comparison, the table also gives $r_u(\textrm{reg})$, which is $B(P_k)$ when $P_k$ is a regular, unit-area k-gon. In all cases, $r_u(\textrm{reg}) > r_u(\textrm{num})$, leading us to conjecture that the maximal value of $r_u$ is $r_u(\textrm{reg})$.

k	3	4	5	6	7	$\infty$ (disc)
$r_u(\textrm{num})$	1.05298	1.08161	1.09583	1.103984	1.107194	–
$r_u(\textrm{reg})$	1.05306	1.08215	1.09821	1.107739	1.113778	$32/(9\pi ) = 1.131768$

It is natural to think that there may be points other than the centroid $\gamma (\Omega )$ which more accurately predict $\max _{\Omega }z$. In fact, we may ask whether there are points $M \in \Omega$ with the stronger property that if, for instance,

$$\begin{aligned} z(V_{i_1}) \le z(V_{i_2}) \le \ldots \le z(V_{i_n}) \end{aligned}$$

(3.38)

then

$$\begin{aligned} |M -V_{i_1}| \le |M -V_{i_2}| \le \ldots \le |M - V_{i_n}|. \end{aligned}$$

(3.39)

We call such M ‘magic points’, and discuss them further in the next section.

3.5 Magic Points

We start by proving

Proposition 3.20

Let T be an arbitrary triangle with centroid $\gamma$. Then $\gamma$ is a magic point for T.

Fig. 7

Left: the triangle T used in the proof of Proposition 3.20. Right: The region QRS in which inequalities (3.41) hold

Proof

Let the vertices of T be $V_1$, $V_2$, $V_3$; the angles at $V_1$, $V_2$, $V_3$ be $\alpha$, $\beta$, $\gamma$ respectively; and $|V_1-V_2| = c$, $|V_3 - V_1| = b$. See Fig. 7. Setting $V_1 = (0, 0)$ and $V_2 = (c, 0)$ we then have that $V_3 = (b\cos \alpha , b\sin \alpha )$. By definition the centroid of T is $\gamma = \frac{1}{3}(V_1 + V_2 + V_3) = \frac{1}{3}(b\cos \alpha + c, b\sin \alpha )$. Defining $d_i^2:= |\gamma - V_i|^2$, $i = 1, 2, 3$, we have

$$\begin{aligned}{} & {} d_1^2 = \frac{c^2 + 2bc \cos \alpha + b^2}{9},\\{} & {} \quad d_2^2 = \frac{4c^2 - 4bc \cos \alpha + b^2}{9},\quad d_3^2 = \frac{c^2 - 4bc \cos \alpha + 4b^2}{9}. \end{aligned}$$

Using the sine rule, we have that $c = b\sin \gamma /\sin \beta$, and the fact that $\alpha + \beta + \gamma = \pi$ gives $\cos \alpha = -\cos (\beta + \gamma )$. Together, these give

$$\begin{aligned} d_1^2&= q\left( \sin ^2\gamma - 2\cos (\beta +\gamma )\sin \beta \sin \gamma +\sin ^2\beta \right) ,\\ d_2^2&= q\left( 4\sin ^2\gamma + 4\cos (\beta +\gamma )\sin \beta \sin \gamma +\sin ^2\beta \right) ,\\ d_3^2&= q\left( \sin ^2\gamma + 4\cos (\beta +\gamma )\sin \beta \sin \gamma +4\sin ^2\beta \right) , \end{aligned}$$

where $q = b^2/(9\sin ^2\beta ) > 0$. Now letting $\Delta _{ij} = (d_i^2 - d_j^2)/q$, we have, from the above, after simplification, that

$$\begin{aligned} \Delta _{12} = -3\sin (2\beta +\gamma )\sin \gamma ,\quad \Delta _{13} = -3\sin (\beta +2\gamma )\sin \beta ,\quad \Delta _{23} = 3(\sin ^2\gamma - \sin ^2\beta ). \end{aligned}$$

(3.40)

Re-labelling the vertices if necessary, we can arrange that $\gamma> \beta > \alpha$, and, since $\alpha , \beta , \gamma$ are the internal angles of a triangle, we have

$$\begin{aligned} \pi> \gamma>\beta> \pi - \beta - \gamma > 0. \end{aligned}$$

(3.41)

Now, by Proposition 2.1, we have, from (3.41), that $\zeta (V_1)> \zeta (V_2) > \zeta (V_3)$, and so $\gamma$ is a magic point for T provided that (3.41) implies that $\Delta _{12}$, $\Delta _{13}$ and $\Delta _{23}$ are all positive.

To prove this, consider the subset $\Sigma := [0, \pi ]\times [0, \pi ]$ of the $\beta$–$\gamma$-plane, in particular, the subset, $\tau$, of $\Sigma$ for which (3.41) holds. See Fig. 7. Then $\tau$ is the intersection of $\Sigma$ and the three half-planes $\gamma > \beta$, $\gamma < \pi - \beta$ and $\gamma > \pi - 2\beta$. Hence, $\tau$ is a triangle, whose vertices are $Q = (\pi /3, \pi /3)$, $R = (\pi /2, \pi /2)$ and $S = (0, \pi )$. Defining $f_1(\beta , \gamma ):= 2\beta + \gamma$ and $f_2(\beta , \gamma ):= \beta + 2\gamma$, we seek the maximum and minimum values of $f_1(\beta , \gamma ), f_2(\beta , \gamma )$ for $(\beta , \gamma )\in \tau$. Since $\nabla f_1 = (2, 1)$ and $\nabla f_2 = (1, 2)$ are both constant, the extrema of $f_1, f_2$ occur at the vertices of $\tau$. Computing directly, we have $f_1(Q) = f_2(Q) = \pi$, $f_1(R) = f_2(R) = 3\pi /2$ and $f_1(S) = \pi$, $f_2(S) = 2\pi$. Hence

$$\begin{aligned} \frac{3\pi }{2}> 2\beta + \gamma> \pi \qquad \text{ and }\qquad 2\pi> \beta + 2\gamma > \pi \end{aligned}$$

and so $\sin (2\beta + \gamma ) < 0$, $\sin (\beta + 2\gamma ) < 0$. Clearly, from (3.41), $\sin \beta$ and $\sin \gamma$ are both positive. Hence, in (3.40), we have that $\Delta _{12} >0$ and $\Delta _{13} > 0$.

Finally, we need to show that $\Delta _{23} = \sin ^2\gamma -\sin ^2\beta > 0$ for $(\beta , \gamma )\in \tau$. Restriction to $\tau$ forces $\beta \in (0, \pi /2)$ and $\gamma \in (\pi /3, \pi )$, so $\sin \beta$ and $\sin \gamma$ are both positive; thus, we need to show that in $\tau$, $\sin \gamma > \sin \beta$. Now, for $\beta \in (0, \pi /2)$, $\sin \gamma > \sin \beta$ provided that $\pi -\beta> \gamma > \beta$; and $\{(\beta , \gamma ): \beta \in (0, \pi /2), \pi -\beta> \gamma > \beta \}\cap \tau = \tau$. Hence, finally, $\Delta _{23} >0$ and $\gamma$ is a magic point for T. $\square$

In order to investigate k-gons with $k > 3$, we carried out computations whose results are summarised in the following table.

k	4	5	6	7	8
$N_c$	100,000	50,000	50,000	50,000	50,000
$N_0$	0	20	193	681	1724

Let us define the magic polygon of a k-gon $P_k$ as the set of magic points for the polygon which lie inside $P_k$. A code was written to compute the magic polygon of a given k-gon directly, by finding the intersection of the appropriate half-planes. We generated $N_c$ random k-gons, with $k = 4, 5, 6, 7, 8$, and the number for which no magic polygon exists, $N_0$, was recorded. (Note that the algorithm necessarily used finite precision arithmetic, 12 significant figures here, so would fail to find a magic polygon if it were sufficiently small.) For $k>4$, we conclude that k-gons which have no magic polygon are rare, though less so as k increases.

Fig. 8

a A pentagon, $P_{nm}$, with no magic polygon. In b–d, we show the stages of the construction of the magic polygon for $P_{nm}$, which is empty

We now give an example of a pentagon, $P_{nm}$, with no magic polygon:

$$\begin{aligned} P_{nm} = (V_1, \ldots , V_5){} & {} = ((-1.1, -0.15), (-0.59, -0.25), \\{} & {} \qquad (0.61, -0.33), (0.74, 0), (0.31, 0.71)), \end{aligned}$$

which is shown in Fig. 8a, along with its ten perpendicular bisectors (dashed lines). We compute that for $P_{nm}$, $z(V_1)> z(V_3)> z(V_5)> z(V_4) > z(V_2)$. In Fig. 8b, the part of $P_{nm}$ to the right—since $z(V_1) > z(V_2)$—of $PB(V_1, V_2)$, outlined with a thick line, is shown, and in (c) is shown the part to the left—since $z(V_3) > z(V_2)$—of $PB(V_2, V_3)$. In (d), what remains in (c) is intersected with the upper (since $z(V_3) > z(V_4)$) half-plane bounded by $PB(V_3, V_4)$, leaving the triangle shown. However, the next inequality, $z(V_5) > z(V_4)$, requires the intersection of this triangle with the lower half-plane bounded by $PB(V_4, V_5)$, shown as a dashed line in (d). This is empty: hence, $P_{nm}$ has no magic points.

By construction, we have therefore proved that pentagons exist with no magic points. On the evidence presented in the table above, we further conjecture that

Conjecture 3.21

All quadrilaterals have at least one magic point, but there exist k-gons for all $k\ge 5$ with no magic points.

There exist, of course, quadrilaterals with a single magic point (for example, squares and rectangles), but it is interesting to conjecture that all quadrilaterals have magic points: the ‘intersection to nothing’ construction shown above for $P_{nm}$ appears to be impossible for quadrilaterals.