How support lines touch an arc

TL;DR

This paper proves a geometric property of simple polygonal arcs, showing they can have at most two pairs of support lines with a specific angular relationship, revealing a new structural constraint.

Contribution

The paper introduces a novel theorem about the maximum number of support line pairs for polygonal arcs with a given angle difference.

Findings

Each simple polygonal arc has at most two such support line pairs.

Support line pairs are characterized by specific points on the arc.

The result constrains the geometric configurations of polygonal arcs.

Abstract

We prove that each simple polygonal arc {\gamma} attains at most two pairs of support lines of given angle difference such that each pair has s1 < s2 < s3 that {\gamma}(s1) and {\gamma}(s3) are on one such line and {\gamma}(s2) is on the other line.

Full text

Department of Mathematics and Computer Science,

Faculty of Science, Chulalongkorn University

1. Interesting way to look at support lines

From [WW], to show that a compact set $K$ can cover every unit arc, it suffices to show that $K$ can cover every simple polygonal unit arc. In the work by Coulton and Movshovich [CM], they prove that for a simple polygonal arc $\gamma$, there is a pair of parallel support lines with points $A$, $B$ and $C$ appear on $\gamma$ in this parametric order such that $A$ and $C$ are one such line and $B$ is on the other line. In this work, we will generalize this result to a pair of lines with any given angle different. In addition, we show that for the parallel case, the pair of support lines is unique.

We first let $\gamma$ be a simple polygonal unit arc parametrized by arc length and suppose that $\gamma$ is not straight. Let $P_{1},...,P_{n}$ be the corner points of its hull which appear in this parametric order with parameters $t_{1},...,t_{n}$. We first consider the multi-valued function $T$ that tells when (in which parameter) a support line touches $\gamma$ as follows. For convenient, we write $L_{\theta}$ for the support line of angle $\theta$ which is the line containing the ray making angle $\theta$ to X-axis and having $\gamma$ on its left side. For example, $L_{0}$ is the horizontal support line under $\gamma$. For each $\theta$, let $T(\theta)=\{\min_{\gamma(s)\in L_{\theta}}s,\max_{\gamma(s)\in L_{\theta}}s\}$. It is clear that $T$ is periodic of period $2\pi$ and rotating $\gamma$ for angle $\theta$ counter clockwise shifts the graph of $T$ to the right for $\theta$. Without loss of generality, we may rotate $\gamma$ so that $P_{1}$ and $P_{2}$ are on $L_{0}$. Hence, accordingly $T$ minimizes at 0 with value $t_{1}$.

Now let us consider the properties of the graph of $T$.

Lemma 1**.**

The graph of $T$ is composed of horizontal segments at level $t_{1},...,t_{n}$.

Proof.

A corner of the hull is corresponding to a horizontal segment. A line segment on the boundary of the hull is corresponding to a vertical 2-valued jump. ∎

Hence $T$ is a step function with finitely many 2-valued jumps. The width of each step is $\pi-\theta$ where $\theta$ is the interior angle of the hull at the corresponding corner of the hull. The different is called the exterior angle. Thus each width is not zero. Let $\delta_{1}$ and $\delta_{n}$ be the widths corresponding to corners $P_{1}$ and $P_{n}$.

Lemma 2**.**

For each $s\in[0,1]$, the horizontal cross section $T^{-1}(s)$ is either empty or a closed interval of length less than $\pi$.

Proof.

If not empty, the cross section is connected, not broken, as follows. Suppose $\theta_{1}$ and $\theta_{2}$ are in the section with $\theta_{1}<\theta_{2}$. Then $L_{\theta_{1}}$ and $L_{\theta_{2}}$ touches $P=\gamma(s)$. Hence $P$ is a corner of the hull. Therefore $[\theta_{1},\theta_{2}]$ is a subset of the section. Let $\theta$ be the exterior angle of the hull at $P$. Thus $0<\theta<\pi$. Since $\gamma$ is not straight, the 2 angles must be at most $\theta$ apart. Therefore the different is less than $\pi$. ∎

Clearly $T$ is not constant. Furthermore, over $[0,2\pi]$, $T$ is initially monotone increasing from $t_{1}$ and finally monotone decreasing back to $t_{1}$. The next lemma shows that $T$ is not zigzag.

Lemma 3**.**

$T$* is monotone increasing and then monotone decreasing on $[0,2\pi]$.*

Proof.

Suppose the contrary to get a contradiction that the graph is zigzag. From the feature of the graph as in Lemma 1, there exist $0<\theta_{1}<\theta_{2}<\theta_{3}<2\pi$ and $s_{1},s_{2},s_{3}$ that $\{s_{i}\}=T(\theta_{i})$ and $t_{1}<s_{1}>s_{2}<s_{3}$ as illustrated by Figure 1.

Firstly we have $t_{1}\leq s_{2}<s_{1}$. Next we will show that $s_{2}>t_{1}$. Suppose that $s_{2}=t_{1}$. Since $\gamma$ is not on a line, $\gamma(t_{1})=P_{1}$ is a corner of the hull touched by $L_{0}$ and $L_{\theta_{2}}$. Hence $T$ is a constant over $[0,\theta_{2}]\ni\theta_{1}$. Then $T(\theta_{1})=\{t_{1}\}$, a contradiction. Therefore $t_{1}<s_{2}<s_{1}$. Since $\theta_{1}<\theta_{3}$ and $\gamma$ is not straight, by Lemma 2, we have $s_{1}\neq s_{3}$. Now we have either $s_{2}<s_{1}<s_{3}$ or $s_{2}<s_{3}<s_{1}$.

In both cases, the subarcs $\gamma(t_{1})\gamma(s_{2})$ and $\gamma(s_{1})\gamma(s_{3})$ intersect (see Figure 2), a contradiction. Therefore $T$ must be simply increasing and then decreasing, not zigzag, on $[0,2\pi]$. ∎

2. Special pairs of support lines

Mainly, we wish to find a pair of support lines with prescribed angle difference that touch $\gamma(s_{1})$ and $\gamma(s_{3})$ by one line and $\gamma(s_{2})$ by the other line such that $s_{1}<s_{2}<s_{3}$ (see Figure 3). More specifically, for a given $\delta<2\pi$, we find $\theta_{1}$ and $\theta_{2}$ such that $|\theta_{1}-\theta_{2}|$ is $\delta$ or $2\pi-\delta$ with $T(\theta_{1})\supseteq\{s_{1},s_{3}\}$ and $T(\theta_{2})\ni s_{2}$ with $s_{1}<s_{2}<s_{3}$.

We will show the existence of this pair of support lines with triple points on them.

Before we deal with the complicate multi-valued function $T$, we will practice on similar functions and get similar results. First we try on continuous functions.

Lemma 4**.**

Let $f:[0,2\pi]\to[0,1]$ be a continuous function with $f(0)=f(2\pi)=0$ and $f(c)=1$ for some $c$ in $(0,2\pi)$. Suppose $f$ is strictly increasing on $[0,c]$ and is strictly decreasing on $[c,2\pi]$ and $0<\delta<2\pi$. Then there exists a unique $x$ such that $f(x)=f(x+\delta)$.

Proof.

Our plan is to find $y$ that $f^{-1}(y)$ contains $x$ and $x+\delta$. To get such $x$ and $y$, we will use inverses of restrictions of $f$ as follows. First note that $f$ is 1-1 and onto on $[0,c]$ and on $[c,2\pi]$. Let $D=(f|_{[c,2\pi]})^{-1}-(f|_{[0,c]})^{-1}$. Since $D$ is strictly decreasing and continuous on $[0,1]$ starting from $2\pi$ down to 0, there is a unique $y$ that $D(y)=\delta$. Equivalently there is a unique $x$ such that $f(x)=y=f(x+\delta)$. ∎

Now we go back to the complicate function $T$. Note that $\delta_{1}$ and $\delta_{n}$ are the exterior angles of the hull at $P_{1}$ and $P_{n}$. From the graph, $\delta_{1}$ and $\delta_{n}$ are the widths of the minimum and maximum sets. From Lemma 2, we have $\delta_{1},\delta_{n}<\pi$. Now we define at each $\theta$, the interval $I_{\theta}$ to be the closed interval $[\min T(\theta),\max T(\theta)]$ (illustratively and correspondingly considered as the vertical segment $\{\theta\}\times I_{\theta}$ over $x=\theta$ in the graph.

Theorem 5**.**

If $\delta_{n}\leq\delta<2\pi$, there exists a unique pair of support lines of angle difference $\delta$ with triple points on them.

Proof.

Our plan is to scan down from the top of the graph over the interval $[0,2\pi]$. Since $\delta\geq\delta_{n}$, we may find $s$ together with $\theta$ such that $s\in I_{\theta}$ and $s\in I_{\theta+\delta}$ (see Figure 4). There the mountain-like graph is $\delta$ wide on level $s$. If $\delta=\delta_{n}$, we have $T^{-1}(t_{n})$ in the form $[\alpha,\alpha+\delta_{n}]$ together with the arc in the situation as illustrated by Figure 5.

Hence we have lines $L_{\alpha}$ and $L_{\alpha+\delta_{n}}$ and $r<t<t_{n}$ such that $\gamma(t)$ is on such line and $\gamma(r)$ and $\gamma(t_{n})=P_{n}$ are on the other line. Now we are in the case $\delta>\delta_{n}$. First we will fill up the gaps where $T$ jumps by defining $T^{|}(\theta)=I(\theta)$. Note that for intervals $I$ and $J$, the substraction $I-J$ is simply $\{i-j|i\in I$ and $j\in J\}$. Let $\theta_{n}$ be in $T^{-1}(t_{n})$. The filled, multi-valued function $T^{|}$ is onto and increasing on $[0,\theta_{n}]$ and is onto and decreasing on $[\theta_{n},2\pi]$ (looks like a step pyramid). Hence the inverses of both restrictions are multi-valued functions that are increasing and decreasing respectively. Let $D=(T^{|}|_{[\theta_{n},2\pi]})^{-1}-(T^{|}|_{[0,\theta_{n}]})^{-1}$. Note that $D$ is a stepping down function with every step filled. Now, since $D$ is decreasing (together with the single-valued functions $\max D$ and $\min D$), there is $s$ that $D(s)$ contains $\delta$. Equivalently there is $\theta$ that $s\in T^{|}(\theta)\cap T^{|}(\theta+\delta)=I(\theta)\cap I(\theta+\delta)$. Suppose both $I(\theta)$ and $I(\theta+\delta)$ degenerate. Then $T(\theta)=T(\theta+\delta)=\{s\}$. Since $\delta>\delta_{n}$, we have $s<t_{n}$. By Lemma 2, the interval $[\theta,\theta+\delta]$ is a subset of $T^{-1}(s)$ and contains $\theta_{n}$ where $T$ takes value $t_{n}>s$, a contradiction. Hence $I(\theta)$ or $I(\theta+\delta)$ does not degenerate. Then one endpoint of such interval is in the other nondegenerated interval. Suppose for the first case that $\min I(\theta)\in I(\theta+\delta)$. Let $s_{1}=\min I(\theta+\delta)$, $s_{2}=\min I(\theta)$ and $s_{3}=\max I(\theta+\delta)$. Thus $s_{1}\leq s_{2}\leq s_{3}$ and $s_{1}<s_{3}$. We have $\gamma(s_{2})$ on $L_{\theta}$ and $\gamma(s_{1})$, $\gamma(s_{3})$ on $L_{\theta+\delta}$. If $s_{1}=s_{2}$, $T^{-1}(s_{1})$ is at least $\delta$ wide, a contradiction. Hence $s_{1}<s_{2}<s_{3}$. Next we will show the uniqueness of $\theta$. Suppose $L_{\theta^{\prime}}$ and $L_{\theta^{\prime}+\delta}$ have such triple points on them with $\theta^{\prime}\neq\theta$. Hence $I_{\theta^{\prime}}\cap I_{\theta^{\prime}+\delta}$ contains some $s^{\prime}$. We may assume $\theta<\theta^{\prime}$. Since $\theta<\theta^{\prime}\leq\max T^{-1}(t_{n})$, $T(\theta)\leq T(\theta^{\prime})$. Similarly, since $\min T^{-1}(t_{n})\leq\theta+\delta<\theta^{\prime}+\delta$, $T(\theta+\delta)\geq T(\theta^{\prime}+\delta)$. We must have $s=s^{\prime}$. Thus $T$ is constant over $[\theta,\theta^{\prime}]$ and over $[\theta+\delta,\theta^{\prime}+\delta]$. By Lemma 2, $T^{-1}(t_{n})$ is longer than $\delta$, a contradiction. The other cases can be treated similarly. ∎

Similarly we have the following theorem.

Theorem 6**.**

If $\delta_{1}\leq\delta<2\pi$, there exist a unique pair of support lines of angle difference $2\pi-\delta$ with triple points on them.

Proof.

Now we scan up from the bottom of the graph over the interval $(0,4\pi)$. Precisely, as the graph looks like “M”, look at the middle part. More precisely, look over the interval $[\theta,\theta+2\pi]$ where $T$ attains maximum at $\theta$. ∎

Corollary 7**.**

Both previous theorems give different pairs of support line for $\delta\neq\pi$. But when $\delta=\pi$, the 2 pairs are identical.

Proof.

Clear. ∎