The $\Lambda$-property of a simple arc

TL;DR

This paper presents a new proof for a notable property of simple polygonal arcs in the plane and explores several generalizations of this property.

Contribution

It offers a novel proof of Coulton and Movshovich's property and extends the concept to broader classes of arcs.

Findings

New proof of the $\Lambda$-property for simple arcs

Generalizations of the property to other arc classes

Enhanced understanding of polygonal arc characteristics

Abstract

In 2006 P. Coulton and Y. Movshovich established an unfamilar but note-worthy general property of simple, polygonal, open arcs in the plane. We give a new and quite different proof of this property, and we consider a few generalizations.

Tables

Table 1. Table 1: Data for the examplar π 0 subscript 𝜋 0 \pi_{0} .

Tilts Spans ${\tau }_0=73.76{}^{\circ }$ ${\tau }_2=25.70{}^{\circ }$ ${\tau }_4=-23.90{}^{\circ }$ ${\tau }_6=-61.01{}^{\circ }$ ${\tau }_1=-69,98{}^{\circ }$ ${\tau }_3=6.34{}^{\circ }$ ${\tau }_5=25.07{}^{\circ }$ ${\tau }_7=66.82{}^{\circ }$ ${\delta }_1=-48.06{}^{\circ }$ ${\delta }_3=-49.60{}^{\circ }$ ${\delta }_5=-37.10{}^{\circ }$ ${\delta }_2=76.32{}^{\circ }$ ${\delta }_4=18.73{}^{\circ }$ ${\delta }_6=41.75{}^{\circ }$

Equations

$$\tau_{j}=\left\{\begin{tabular}[]{ll}$\measuredangle\left(\xi,HX\right)$&\quad$j=0,$\\ $\measuredangle\left(\xi,B_{j}\right)$&\quad$j=1,2,\dots,J,$\\ $\measuredangle\left(\xi\mathfrak{,}YT\right)$&\quad$j=J+1,$\end{tabular}\right.$$

$$\left\{\begin{tabular}[]{ll}$0{{}^{\circ}}<\tau_{J+1}<180$&\quad when $J$ is even,\vskip 6.0pt plus 2.0pt minus 2.0pt\\ $-180{{}^{\circ}}<\tau_{J+1}<0{{}^{\circ}}$&\quad when $J$ is odd.\end{tabular}\right.$$

$$\tau_{0}>\tau_{2}>\tau_{4}>\cdots>\left\{\begin{tabular}[]{ll}$\tau_{{}_{J}}$&\quad if $J$ is even,\\ $\tau_{{}_{J+1}}$&\quad if $J$ is odd,\end{tabular}\right.$$

$$\mathfrak{\delta}_{j}=\left\{\begin{tabular}[]{ll}$\tau_{j+1}-\tau_{j-1}=\measuredangle(B_{j-1},B_{j+1})$&\quad when $j$ is even,\\ $\tau_{j+1}-\tau_{j-1}=\measuredangle(B_{j-1},B_{j+1})$&\quad when $j$ is odd and $j<J,$\\ $\tau_{J+1}-\tau_{J-1}=\measuredangle\left(B_{J-2},B_{J}\right)$&\quad when $J$ is odd.\end{tabular}\right.$$

$$L_{j}=\left\{\begin{tabular}[]{ll}$\left\{\left(x,y\right):0\leq x\,<\left|\delta_{1}\right|,-180{{}^{\circ}}<y\leq 180{{}^{\circ}}\right\}$&\quad$j=1,\vskip 3.0pt plus 1.0pt minus 1.0pt$\\ $\left\{\left(x,y\right):\sum\nolimits_{1}^{j-1}\left|\delta_{i}\right|\leq x<\sum\nolimits_{1}^{j}\left|\delta_{i}\right|,\text{ }-180{{}^{\circ}}<y\leq 180{{}^{\circ}}\right\}$&\quad$j=2,3,\ldots,J.$\end{tabular}\right.$$

$$T_{j}=\left\{\begin{tabular}[]{ll}$\left(0,\tau_{1}\right)$&\quad$j=1,\vskip 3.0pt plus 1.0pt minus 1.0pt$\\ $\left(\sum\nolimits_{1}^{j-1}\left|\delta_{j}\right|,\tau_{j}\right)$&\quad$j=2,3,\ldots,J+1.$\end{tabular}\right..$$

$$\left.\begin{tabular}[]{l}$\Upsilon=\langle T_{0}T_{2}T_{4}\cdots T_{J}\rangle$\\ $\Phi=\langle T_{1}T_{3}T_{5}\cdots T_{J+1}\rangle$\end{tabular}\right\}$$

$${\phi }_L$$

$${\phi }_R$$

$${\phi }_L$$

$$\left|{\phi }_R\right|$$

Full text

AMS 2010 subject classification:  Primary: 51N05; secondary: 51M99.

Keywords and phrases:  simple arcs, open arcs, polygonal arcs, support lines for arcs, Coulton-Movshovich $\boldsymbol{\Lambda}$ property.

We give a new proof of a property of open simple arcs proved for polygonal arcs in 2006 by P. Coulton and Y. Movshovich [1, Th. 5.1]:

They called this the ${\Lambda}$-property of the arc.

In the concluding section we extend this theorem to the case in which the support lines are to form a given angle, and we consider a few related results.

1 The parallel case.

Throughout, by an open simple arc $\gamma$ we mean the one-to-one continuous rectifiable image of the real interval $[0,1]$ with $\gamma(0)\neq\gamma(1).$  For simplicity of language we do not distinguish notationally between the mapping function $\gamma$ and the pointset that is its range.

Our proof of Theorem 1 begins with a new proof for the case of polygonal arcs.  The result for arbitrary arcs then follows from classical approximation reasoning.

1.1 Initial considerations.

Let $\pi=\langle p_{1}p_{2}\cdots p_{P}\rangle$ be an open, simple, oriented, polygonal arc, which we suppose is not convex and, in particular, is not a line segment.  The points $p_{i}$ are the nodes of $\pi$.  The closed convex hull $\mathfrak{K}$ of $\pi$ is a polygon.  Let $\partial\mathfrak{K}$ be its boundary, and $\mathcal{V}$ its set of vertices (a vertex of $\mathfrak{K}$ is a point of $\mathfrak{K}$ that does not lie between any two points of $\mathfrak{K}$).  If $\mathfrak{K}$ is a triangle, then one properly chosen side and the parallel line through the opposite vertex, together with the three suitably labeled vertices, satisfy the conditions of the Theorem.  Consequently we may assume from the start that $\mathfrak{K}$ has $n\geq 4$ vertices.

The segments that connect consecutive vertices on $\partial\mathfrak{K}$ are the edges of $\mathfrak{K}.$  Each vertex of $\mathfrak{K}$ is a node of $\pi,$ but not necessarily conversely (Figure 2).

A support line of $\mathfrak{K}$ can touch $\mathfrak{K}$ only at a vertex or along an edge.

1.1.1 Polygonal paths in $\mathfrak{K.}$

We are interested in polygonal paths that capture the convexity properties of $\mathfrak{K}$.

**Definition **

A polygonal path $\varpi=\langle w_{1}w_{2}\cdots w_{n}\rangle$ is admissible if (1) it lies in $\mathfrak{K},$ (2) it is simple, and (3) the set $\left\{w_{1},w_{2},\cdots,w_{n}\right\}$ of its nodes is equal to the vertex set $\mathcal{V}$ of $\mathfrak{K}$.

To generate such a path, select a vertex of $\mathfrak{K}$ to be the initial node.  Once any node but the penultimate one has been located at a vertex of $\mathfrak{K}$, there are precisely two vertices available for the next node, because the path must be simple and it must visit each vertex of $\mathfrak{K}$ exactly once.  When the penultimate node has been placed, just one vertex remains for the last node.  Consequently:

Lemma 2

A strictly convex $n$-gon $\mathfrak{K}$ has $n2^{n-2}$ admissible polygonal paths each of which has $n$ nodes, and $\mathfrak{K}$ is the convex hull of each.

1.1.2 The guide path $\varpi$ for $\pi.$

Among all admissible paths there is just one that visits the vertices of $\mathfrak{K}$ in the same sequence that they are visited by the given arc $\pi.$  We call it the guide path of $\pi,$ and we reserve for it the notation $\varpi=\langle w_{1}w_{2}\cdots w_{n}\rangle$.  The guide path of $\pi$ serves as a “canonical form” for $\pi.$  It is open and simple, and it clearly is the shortest arc that spans the convex hull of $\pi$ and touches its vertices in the same sequential order as does $\pi.$

We call $w_{1}$ the head and $w_{n}$ the tail of $\mathfrak{K}$, and recalling that $n\geq 4$ we introduce the notation $H=w_{1},$ $X=w_{2},$ $Y=w_{n-1},$ and $T=w_{n}.$

We call the oriented line segment $HT$ the axis $\xi$ of $\mathfrak{K.}$  If the axis $\xi$ lies on the boundary of $\mathfrak{K,}$ then $H$ and $T$ are adjacent vertices, and the choices $m$ for the line $HT,$ $n$ the support line of $\mathfrak{K}$ parallel to $m,$ $u=H,$ $w=T,$ and $v$ a vertex in which $n$ meets $\mathfrak{K}$ show that $\pi$ satisfies the theorem; so we examine the case in which the axis $\xi$ meets the interior of $\mathfrak{K.}$  Then the points $H$ and $T$ divide the boundary of $\mathfrak{K}$ into two convex arcs, one of which passes through the vertex $X$.  We call that arc the *upper boundary *of $\mathfrak{K}$ and the other arc the *lower boundary.  *The upper and lower boundary arcs lie on opposite sides of the axis of $\mathfrak{K.}$  We orient each edge $[v_{i},v_{i+1}]$ of $\partial\mathfrak{K}$ in the sense from $H$ to $T.$

The oriented segments $[w_{i},w_{i+1}]$ are the links of $\varpi.$  These links are of two kinds:  those that lie on $\partial\mathfrak{K}$ and those that cross $\mathrm{int}(\mathfrak{K})$ and meet the axis $\xi\mathfrak{.}$  If a link of the guide path $\varpi$ lies on the boundary $\partial\mathfrak{K},$ we call it an edge link.  It coincides with an edge of $\mathfrak{K}$.  A link of $\varpi$ that is not an edge link is a crossing link.

Exemplar.

It is our intention to follow the general argument using the specific arc pictured in Figure 2(a) as an exemplar, and for this purpose we adopt the notations $\pi_{0}$ for this particular arc and $\mathfrak{K}_{0}$ for its convex hull, pictured in Figure 2(b).  Evidently $\pi_{0}$ has $P=60$ nodes, and its convex hull $\mathfrak{K}_{0}$ has $16$ vertices.  The guide path $\varpi_{0}$ for the exemplar $\pi_{0}$ is shown bold in Figure 3.

1.1.3 Locales of $\pi.$

The guide path $\varpi$ partitions the convex hull $\mathfrak{K}$ into $J$ closed convex subregions $\mathfrak{L}_{j}$ called *locales, *indexed $\mathfrak{L}_{1},$ $\mathfrak{L}_{2},$ $\dots$ $\mathfrak{L}_{J}$ in order from $H$ to $T$  Each locale $\mathfrak{L}_{\ell}$ has an edge link $B_{j}$ as its base, one or two crossing links as its sides, and a chain $C_{j}$ of edge links as its cap.  The cap $C_{1}$ lies on the upper boundary of $\mathfrak{K}$ and the base $B_{1}$ on the lower boundary, and for increasing index the locations alternate, with caps on the upper boundary and bases on the lower when $j$ is even, and caps on the lower and bases on the upper when $j$ is odd.  The situation at $T,$ when $j=J,$ depends on the parity of $J.$

Exemplar.

Figure 3 shows the guide path $\varpi_{0}$ of the exemplar $\pi_{0}$ and its associated six locales.  Note that like $C_{4},$ the cap of a locale can be a single point.

1.1.4 The angles of $\mathfrak{K}$.

Two important angles, tilt and *span, are associated with each locale of $\varpi.$  *In describing them we write $\measuredangle\left(r_{u},r_{v}\right)$ for the directed angle from a ray $r_{u}$ to the ray $r_{v}$ in the range $(-180{{}^{\circ}},180{{}^{\circ}}],$ positive in the counterclockwise sense and negative in the clockwise sense.

Tilt.

For each index $j,$ $0\leq j\leq$ $J+1,$ the tilt $\tau_{j}$ is defined to be

[TABLE]

where $\xi$ is the axis, oriented from $H$ to $T.$  (The situation is illustrated in Figure 4, in which the terminal side of each tilt angle $\measuredangle\left(\xi\mathfrak{,\tau}_{j}\right)$ is marked $\tau_{j}$.)  The tilt of $\mathfrak{L}_{j}$ measures the directed angle of its base with respect to the horizontal axis $\xi$.

Lemma 3

Initial inequalities for $\tau_{0},$ $\tau_{1},$ and $\tau_{J+1}$ are: $0{{}^{\circ}}<\tau_{0}<180{{}^{\circ}},\ -180<\tau_{1}<0{{}^{\circ}},$ and

[TABLE]

The tilts satisfy the monotonicity inequalities

[TABLE]

Proof. The inequalities for $\tau_{0}$ and $\tau_{1}$ are the result of the normalization, and the monotonicity claims are consequences of the convexity of $\mathfrak{K.}$

Span.

The span $\delta_{j}$ of the interior locale $\mathfrak{L}_{j}$ is defined in terms of tilts, as follows.  For $j=1,2,\dots,J$:

[TABLE]

When $j$ is even, $\delta_{j}>0,$ reflecting the fact that the support lines of $\mathfrak{K}$ at points of the cap $C_{j}$ turn counterclockwise across $C_{j}.$  Similarly, when $J$ is odd, $\delta_{j}<0,$ and the support lines of $\mathfrak{K}$ a points of $C_{j}$ turn clockwise across $C_{j}$ (see Figure 5).

For every ray $r$ between $B_{j-1}$ and $B_{j+1}$ there is exactly one line parallel to $r$ that touches the cap $C_{j}$ of $\mathfrak{L}_{j}.$  Consequently, the span of $\mathfrak{L}_{j}$ is the range of the angles of supporting lines that contact $\mathfrak{K}$ on the cap $C_{j}.$

1.2 The schematic diagram.

To analyze the data for $\pi$, we employ a schematic diagram built in the square $[0{{}^{\circ}},360{{}^{\circ}}]\times\left(-180{{}^{\circ}},180{{}^{\circ}}\right)$ in $\mathbb{R}^{2}$.  The units on each axis are degrees.

Let $\delta=|\delta_{1}|+|\delta_{2}|+\cdots+|\delta_{J}|,$ so that $0<\delta\leq 360{{}^{\circ}}.$  On the interval $\left[0,\delta\right]$ on the $x$-axis, mark off successive intervals of lengths $|\delta_{1}|,$ $|\delta_{2}|,$ $\dots,$ $|\delta_{J}|,$ and locate a point on the lateral edges of each rectangle

[TABLE]

by setting

[TABLE]

The rectangular region $L_{j}$ encodes the angle data for the locale $\mathfrak{L}_{j}$:  its width is the span $\left|\delta_{j}\right|$ of $\mathfrak{L}_{j}$ and the point $T_{j}$ on its left edge gives the tilt of its base.

In each rectangle $L_{j}$ we insert a horizontal line segment whose left endpoint is $T_{j}.$  Recalling that the width $\left|\delta_{j}\right|$ is precisely the spacing $\left|\tau_{j-1}-\tau_{j+1}\right|$ (1), join the consecutive even-indexed points $T_{j}$ with line segments having slope $-1$ and the consecutive odd-indexed points $T_{j}$ with line segments having slope $+1,$ as pictured in (Figure 6).  Then the horizontal segment at $y=\tau_{j}$ marks the angle of the base of $\mathfrak{L}_{j},$ and the $\pm 45{{}^{\circ}}$ segment codes the range $\delta_{j}$ of angles of the support lines that touch $\mathfrak{L}_{j}$ on its cap $C_{j}.$

The horizontal and slant line segments assemble themselves into two continuous piecewise linear paths $\Upsilon$ and $\Phi$ that code the contact angles $\vartheta$ of the support lines $\ell_{\vartheta}$ that contact $\mathfrak{K}$ on its upper boundary ($\Upsilon$) and on its lower boundary ($\Phi$), respectively.  More precisely, define

[TABLE]

Recalling that $\tau_{0}$ is positive and $\tau_{1}$ is negative, and $\tau_{J}$ and $\tau_{J+!}$ always have opposite signs, we see that $\Upsilon$ is a continuous, decreasing, piecewise linear function on $\left[0,\delta\right]$ that falls from $T_{0}$ to $T_{J}$ (or to $T_{J+1}$), and $\Phi$ is a continuous, increasing piecewise linear function on $\left[0,\delta\right]$ that rises from $T_{1}$ to $T_{J}$ (or to $T_{J+1}$) (Figure 6).  The fact that these two graphs must meet and cross constitutes the proof of Theorem 1.

We have established the following:

Lemma 4

The piecewise linear functions $\Upsilon$ and $\Phi$ defined by (2) meet and cross at precisely one point $M$ inside a strip $L_{j}$ that represents precisely one locale $\mathfrak{L}_{j}$, and through $M$ there passes exactly one horizontal segment and exactly one slant segment.

Proof. The decreasing function $\Upsilon$ starts at [math] above and ends at $\delta$ below the increasing function $\Phi,$ and the result follows from continuity.

1.3 The proof of Theorem 1 for polygonal arcs.

The point $M$ in which the paths $\Upsilon$ and $\Phi$ meet and cross lies in exactly one rectangle $L_{j}$ representing precisely one locale $\mathfrak{L}_{j},$ and through $M$ pass exactly one horizontal segment and exactly one $\pm 45{{}^{\circ}}$ segment.  Let $m$ be the (extended) base $B_{j}$ of $\mathfrak{L}_{j}.$  There is exactly one line $n$ parallel to $m$ and touches $\mathfrak{K}$ on the cap $C_{j},$ say at the point $v.$  Let $u$ and $w$ be the endpoints of the base segment $B_{j}$ of $\mathfrak{L}_{j}.$  The conditions of Theorem 1 are satisfied by the lines $m$ and $n$ and the vertices $u,$ $v,$ and $w$; and the proof for simple, open, polygonal arcs is completed.

Exemplar.

The drawings involving the exemplar $\pi_{0}$ were made using a CAD program, which provided the angular data assembled in Table 1.

Figure 6(a) shows the schematic diagram for $\pi_{0}$ drawn from the angle data in Table 1.  The crossing point $M$ falls in locale $\mathfrak{L}_{3},$ the line $m$ passes through the vertices $u=w_{7}$ and $w=w_{11}$ and the point $v$ is the vertex $w_{9}$ (Figure 7).

1.4 The extension to rectifiable arcs.

Classical approximation methods (see, for example, [2] and [5]) can be used to extend the proof for open simple polygonal arcs to open simple rectifiable arcs, but a little care must be taken to avoid the confluence of the three points of contact, in which case one can show that the original curve must have been a straight segment.  We leave the details to the interested reader.

1.5 Closed simple arcs.

We say that a simple closed rectifiable arc $\gamma$ has the $\Lambda$ *Property *if it has two parallel support lines $m$ and $n$ with three different points $u,$ $v,$ and $w$ of $\gamma$  with $u$ and $w$ on $m$ and $v$ on $n.$  Recall that a convex set with non-empty interior is called strictly convex if its boundary contains no straight line segments (Valentine [3, p. 94]).  In this regard we have the following general theorem:

Theorem 5

A simple rectifiable arc $\gamma$ has the $\Lambda$ property if and only if its convex hull is not strictly convex.

Proof. If $\gamma$ is open, then it has $\Lambda$ property, and the line segment $[u,w]$ lies on the boundary of its convex hull.  Conversely if the line segment $[u,w]$ lies in the boundary of its convex hull $\mathfrak{K}$, then the support line $n$ parallel to $m=\overleftrightarrow{uw}$ meets $\mathfrak{K}$ at a point $v,$ and the conditions are satisfied.

Suppose $\gamma$ is closed, and let $\mathfrak{K}$ be its convex hull.  If $\gamma$ has the Movshovich property, then again the line segment $[u,w]$ lies in the boundary $\partial\mathfrak{K}$ of the convex hull, so $\mathfrak{K}$ is not strictly convex.  Finally, if $\mathfrak{K}$ is not strictly convex, there are points $u$ and $w$ so that the line segment $[u,w]$ lies in $\partial\mathfrak{K.}$  The line $m=\overleftrightarrow{uw}$ is a support line of $\gamma;$ let $n$ be the support line parallel to $m,$ and suppose that $n$ meets $\partial\mathfrak{K}$ at a point $v.$  Then $m,$ $n,$ $u,$ $v,$ and $w$ have the desired properties.

The uniqueness assertion may fail for simple closed curves, convex or not.

2 The general case.

Similar results hold for support lines required to be at prescribed angles.  As above we begin with an open, simple, polygonal arc $\pi$ and a given positive angle $\varphi,$ and we seek support lines $m$  and $n$ and points $u,$ $v,$ $w$ on $\pi$ so that $\left|\measuredangle\left(m,n\right)\right|=\varphi,$ the points $u$ and $w$ lie on $m,$ the point $v$ lies on $n,$ and $u,$ $v,$ and $w$ fall in the order $u$-$v$-$w$ in the parametric ordering of $\pi.$  Here is the precise result.

Theorem 6

An open simple, polygonal arc $\pi$ (not a line segment) and an angle $\varphi\geq 0$ are given.  Let $K$ and $\varpi$ be the convex hull and guide path of $\pi.$  Partition $K$ into its locales $\mathcal{L}_{1},$ $\mathcal{L}_{2},$ $\dots$ $\mathcal{L}_{J}$, so that the tilts $\tau_{j}$ are known.  Let

[TABLE]

Then $\varphi_{L}>0$ and $\varphi_{R}<0;$ we call the key angles $\varphi_{L}$ and $\varphi_{R}$ the left and right aspect angles of $\pi,$ respectively.

A.

*If $\varphi=0,$ then there is exactly one pair $(m,n)$ of parallel support lines of $\mathfrak{K}$ that has the * $\Lambda$ property.

B.

*If $0<\varphi\leq\min\left\{\varphi_{L},\left|\varphi_{R}\right|\right\},$ then there are exactly two pairs $\left(m,n\right)$ of support lines of $\mathfrak{K}$ with $\left|\measuredangle\left(m,n\right)\right|=\varphi$ having the $\Lambda$ property, and for one the point $m\cap n$ lies to the left of $H$ and for the other the point *$m\cap n$ lies to the right of $H$.

C.

*If $\min\left\{\varphi_{L},\left|\varphi_{R}\right|\right\}<\varphi\leq\max\left\{\varphi_{L},\left|\varphi_{R}\right|\right\},$ then there is exactly one pair $\left(m,n\right)$ of support lines of $\mathfrak{K}$ with $\left|\measuredangle\left(m,n\right)\right|=\varphi$ having the *$\Lambda$ property.

D.

If $\max\left\{\varphi_{L},\left|\varphi_{R}\right|\right\}<\varphi,$ then there are no pairs $\left(m,n\right)$ of support lines that satisfy the desired conditions.

Proof. Part (A) is Theorem 1.  The remaining parts follow immediately from a modification of the schematic diagram (Figure 6).  We insert the graph of the difference function $\Delta=\Upsilon-\Phi$ into that diagram to form an extended schematic diagram (Figure 9).

The graph of $\Delta$ is a line segment with slope $-1$ and endpoints $(0{{}^{\circ}},\vartheta_{L})$ and $(\delta,\vartheta_{R}).$  If the line at ordinate $\varphi$ meets the graph of $\Delta$ at a point in a rectangle $L_{j},$ then the extended base $B_{j}$ of the associated locale $\mathfrak{L}_{j}$ serves as the support line $m$ and the endpoints $u$ and $w$ of $B_{j}$ as two of the three desired points; and the second support line $n$ can be found making the desired angle $\varphi$ with $m$ and contacting the locale $\mathfrak{L}_{j}$ at a point $v$ of its cap $C_{j},$ and evidently $u$-$v$-$w.$ The three remaining cases of the Theorem reflect the fact that the horizontal lines at ordinates $\pm\varphi$ meet the graph of $\Delta$ at a total of two, one, or no points.

Exemplar.

The aspect angles of the exemplar $\pi_{0}$ are

[TABLE]

Hence for any angle $\varphi$ with $0<\left|\varphi\right|\leq$ $127.833{{}^{\circ}}$ there are exactly two pairs of support lines $\left(m,n\right)$ that satisfy the Movshovich conditions, and for any angle $\varphi$ with $127.833{{}^{\circ}}<\varphi\leq$ $143.743{{}^{\circ}}$ there is exactly one such pair of support lines.  For example, for $\varphi=40{{}^{\circ}}$ there are two support pairs for $\pi_{0}$, found using the dashed lines at ordinates $\pm 40{{}^{\circ}}$ in Figure 9(a); and for $\varphi=135{{}^{\circ}}$ there is just one pair of such support lines, found from the lighter dashed line at ordinate $-135{{}^{\circ}}$ in this figure.  These support lines are pictured in Figure 10.

Remark.  In 1970 one of us conjectured that a 30∘ circular sector having unit radius is a cover for the family $\mathcal{F}$ of all arcs of unit length (see [4, p. 358]), i.e., that this circular sector contains an isometric copy of every such arc.  This conjecture has not been resolved, although there have many partial results, mostly unpublished.  If this sector is such a cover, it would be the smallest convex cover for $\mathcal{F}$ currently known.  One reason for our interest in Theorem 6 is the hope that it might cast some light on this conjecture.