an analysis of Mosers worm

This article is based off a previous writeup on the subject. To read that article, click here.

Introduction

#

Moser’s Worm is an important unsolved problem in geometry with a very amusing name. The name comes from an odd hypothetical. Suppose you have an inch-long pet worm, who’s a very restless sleeper. When the worm goes to bed, there’s no way of knowing how it might be twisted or curved. Nonetheless, you want to cover your worm with a little blanket. What’s the smallest blanket you could use, regardless of how the worm is positioned?

Although many unsolved problems in mathematics are known for their incomprehensibility, the standard statement of Moser’s worm only requires requires one definition:

From here on out, I might be a little bit looser. I’ll be certain to eliminate ambiguity whenever necesary, but the point of these notes is mostly to ensure that these concepts are written down somewhere they’re easy to locate later on, without necessarily being up to the hyper-rigorous standards of mathematical publication. This is my website, after all. Without further ado, here’s the statement of the problem:

It’s a shockingly simple question for something that has remained unsolved for so long! Although my background is in mathematics, and I studied differential geometry in my undergrad, this question in particular is one I don’t know a lot about. It’s a bit refreshing to work on something unrelated to what you normally do, so I’m looking forward to it! I should mention that I’m deliberately not looking into the already existing body of work on this problem. This is partly for my own satisfaction of discovering things for myself, and partly to avoid bias. Although I don’t suppose that I could fully crack an infamous problem outside my own specialty, it can still be very invigorating for a field to get a fresh pair of eyes and a fresh perspective.

There’s also the second benefit of writing an article on a subject from an outsider’s perspective, which is that it should be very approachable for the reader. Because I don’t know any of the jargon, techniques, or theorems in this field, I can’t possibly confuse a reader by using them. That said, I will use some basic calculus at a couple points.

Type 1 and Type 2 Curves

#

My goal here is to see what curves we don’t need to worry about. There’s two prongs to this:

1. Show that curve $A$ is always accommodated when curve $B$ is accommodated.
2. Show that when curves $A$ and $B$ are accommodated, curve $C$ is also accommodated.

To that end, we give these definitions:

This entire problem exists in euclidean space. If $d$ is the euclidean distance metric, and $a$ and $b$ are points, then $\ell\langle a,b\rangle$ will be used as a synonym for $d(a,b)$

Immediately there are some lemmas we can form regarding taut curves without any further context. For example, this is the lemma that inspired the choice of the word “taut”:

proof:
Let $X$ be a unit-length curve, and let $x$ be a contiguous section of $X$ that does not touch the border of $[X]$, except at two endpoints, $a$ and $b$. Then let $X’$ be identical to $X$, except that $x$ is replaced with a straight line from $a$ to $b$. The length of $X’$ is necessarily less than 1, although $X’$ has the same convex hull. Finally, let $X’’$ be the unit-length curve that includes all of $X’$, but with an added length at the end. $X’’$ must have a convex hull that includes everything in $[X]$ and more.

Hopefully, this drawing explains why I chose the word "taut" to describe this property. The curve on the left is "taut", in the sense that it could be stretched to coincide with the border of its convex hull, and that extra length could be used to create a larger hull. The light grey area is the same between the two shapes, but the dark grey area is added by using length more efficiently

Here’s another brief and straightforward observation about taut curves.

Let’s start our effort to characterize taut curves with yet another definition

An example of a Type 1 and Type 2 curve.

Type 1 and Type 2 curves can also be thought of in terms of path direction. A path along a Type 1 curve moves entirely either clockwise or counterclockwise across the edge of its convex hull. A path along a Type 2 curve starts moving either clockwise or counterclockwise along the border, but then moves in the opposite direction for a length.

Here’s a brief lemma about Type $n$ curves that will be useful in several results coming up

proof:
Suppose, for the sake of contradiction, that $X_a^b$ does not have zero curvature. Then $\ell X_a^b>\ell\langle a,b\rangle$. Let $X’$ be formed by replacing $X_a^b$ in $X$ with $\langle a,b\rangle$. It is clear that $\ell X’ < X$, and so $X$ is accommodated by $X’$, and is not taut.

I’ve only talked about Type 1 and Type 2 curves so far. Why hasn’t any special attention been given to curves of other types? As it turns out, these curves are never taut.

proof:
To force a contradiction, let $X$ be such a curve. Let $a$, $b$, $c$, $d$, $e$, $f$, and $g$ be points on $X$, appearing in that order, with the added requirements that:

1. $a$, $b$, $c$, $d$, $e$, $f$, and $g$ are all on the boundary of $[X]$.
2. $a$ and $g$ are the endpoints of $X$.
3. $d$ is the point furthest from $\langle a,g \rangle$, on the opposite side from $b$ and $f$
4. $\langle b,c \rangle$ is internal to the convex hull of $X$.
5. $\langle e,f \rangle$ is internal to the convex hull of $X$.

Here’s a drawing to explain.

For ease of upcoming calculations, let $\langle a,g\rangle$ lie on the $x$ axis, with $a$ existing at $0,0$.

Let $\phi$ be the minimal distance from $d$ to $\langle a,g \rangle$. It immediately follows that:

$$\frac{\ell X_b^f}{2}\leq\phi$$

In this case, define $a’$ and $g’$ to be these two points:

$$a’=\left( 0, -\frac{\ell X_b^f}{2}\right)$$

$$g’=\left( \ell\langle a,g\rangle, -\frac{\ell X_b^f}{2}\right)$$

Then we can define $X’$ to be the curve

$$X’ = \langle a’,a\rangle \cup X_a^b \cup \langle b,f\rangle \cup X_f^g \cup \langle g,g’\rangle$$

This is given in the drawing below:

Because $a’$ and $g’$ are both below $d$, the entirety of $X$ must be inside $X’$.

Let’s limit a bit further which curves can be Type 1.

proof:
Let $X$ be a curve as described above, and let $c$ be the point on $X$ that lies furthest from any line perpendicular to $\langle a, b\rangle$. without loss of generality, assume $c$ is closer to $a$ than it is to $b$.

Let $\alpha$ be the line that includes the line segment $\langle a, b\rangle$, and let $c’$ be the point on $\alpha$ closest to $c$. Let $X’$ be a curve identical to $X$, except that the section between $c$ and $a$ is replaced with a straight line between $c$ and $c’$.

This situation is illustrated below:

It is clear that

$$\ell\langle c,c’\rangle < \ell C_c^a$$

and therefore

$$\ell X’<\ell X$$

Therefore, $X’$ has a length less than $X$, and so is a unit curve. The convex hull of $X’$ can accommodate $X$, but is larger. By definition, $X$ is not taut.

Finally, these past few lemmas and a few other ideas are compiled in a theorem that helps to characterize Type 1 curves:

proof:
The fist of these requirements is implied by the definition of a Type 1 curve, and the second is implied by Lemma 3. Therefore, this proof focuses on the third requirement

First, suppose that a set of points as described exists. Without loss of generality, let $f=a$. As before, I’ll include a diagram to clarify what the situation described in the lemma statement is.

Let $X’ = \langle c,a,b\rangle \cup X_b^e \cup \langle e,d\rangle$. As evident from the diagram, the convex hull of $X’$ accommodates the convex hull of $X$. Next, consider the inequality assumed earlier:

$$\begin{align*} \ell \langle a,b\rangle + \ell \langle c,a\rangle + \ell\langle e,d\rangle &\leq \ell X_a^e \\ \ell \langle c,a,b\rangle + \ell\langle e,d\rangle &\leq \ell X_a^e \\ \ell \langle c,a,b\rangle + \ell X_b^e + \ell\langle e,d\rangle &\leq \ell X_a^e + \ell X_b^e \\ \ell X’ &\leq \ell X \end{align*}$$

Therefore, $[X’]$ both accommodates $[X]$ and includes other area as well, but $X’$ is shorter than $X$. By definition, $X$ is not taught.

An “if and only if” proof requires that the statement is proven in both directions, so Theorem 2 will now be proved in the opposite direction. Suppose that such a set of points as ${c, d, e, f}$ does not exist, but that $X_1$ is a Type 1 curve such that $[X_1] \sqsupset X$. Let $c$ and $d$ be the endpoints of $X_1$. An example of what this might look like is shown below:

Next, let $\alpha$ be the line passing through $c$ and $d$. let $X_2$ be the shortest curve with endpoints on $\alpha$ that accommodates $X$. $X_2$ must consist of two lines parallel to $\alpha$, as well as a section of the boundary of $[X]$, which may or may not include $\langle a,b\rangle$. Let $c’$ and $d’$ be the endpoints of $X_2$, and let $e$ and $f$ be the points where $X_2$ meets the convex hull of $X$, as shown below:

Finally, the length of $X_2$ can be reduced even further. Let $d’’$ and $c’’$ be points on $\langle d’,e\rangle$ and $\langle c’,e \rangle$ that are both a distance $\phi$ away from $d’$ and $c’$, with $\phi$ having the largest possible value without $\langle d’’,c’’\rangle$ intersecting the interior of $[X]$.

Let $X_3$ be $X_2$, without the lines from $d’$ to $d’’$ and from $c’$ to $c’’$. If $f$ and $e$ are both endpoints of $X$, then $X_3$ must contain the entirety of $X$. This is impossible, as $\ell X = 1$, so either $e$ or $f$ is not an endpoint of $X$.

Instead, assume (for the sake of contradiction), that neither $f$ nor $e$ are endpoints of $X$. This would require that at least one of the two points is not on a line parallel to $\langle a,b \rangle$.

Therefore, $\lbrace c’’, f, e, d’’\rbrace$ satisfies the requirements of the lemma.

The reason I set out this lemma is because I want to demonstrate that there is a limit on the “height” of Type 1 curves. If $\langle a,b \rangle$ is very short, and $X$ has a point very distant from $\langle a,b \rangle$, it would be easy to find a way to maximize $\ell X_a^e$ and minimize $\ell \langle a,c \rangle$

This theorem serves to characterize type 1 curves fairly precisely. One of the broader goal of this project is to characterize Type 2 curves equally as well.

proof:
The locations of $a$ though $d$ can be clarified by this diagram:

Suppose for the sake of contradiction that $\ell\langle a.c\rangle > \ell\langle b,c\rangle$. Define $X’$ such that

$$X’ = X_b^a \cup \langle a,c\rangle \cup X_c^d$$

Because $\ell\langle a,c\rangle > \ell\langle b,c\rangle$, $\ell X’ < \ell X = 1$. Then let $X’’$ be identical to $X’$, except that an extra bit is added to increase the area of the convex hull while keeping $\ell X’’\leq 1$. By definition, $X$ is not taut.

By symmetry, $X$ is also not taut if $\ell \langle b.d\rangle > \ell \langle b,c\rangle$.

The non-internal sections of a Type 2 curve behave something like Type 1 curves themselves. For example, consider the following Lemma:

proof:
First, for clarity, a diagram of these lines is shown:

First, if $p$ is a point on $X_c^d$ that is on the “wrong” side of $L_1$, then $c$ is an internal point of $[X]$. This is not true by assumption.

The point $p$ must be on the exterior of $[X]$, by assumption, and so cannot be on the wrong side of $\langle c,d\rangle$.

Finally, suppose for the sake of contradiction that $p$ lies on the wrong side of $L_2$. Without loss of generality, let $p$ specifically be the point furthest from $L_2$. Then let $p’$ be the point on the line containing $b$ and $d$ that is closest to $p$. If $X’ = X_a^p \bigcup \langle p, p’\rangle$, then $X’$ is shorter than $X$. Because $\langle p,p’\rangle$ is perpendicular to $\langle p’,d\rangle$, $p’$ must also be further from $L_2$ than any other point in $X$. Therefore, $X’$ accommodates $X$, giving a contradiction.

This lemma comes with a few charming corollaries:

Finally, we can make a little theorem effectively characterizing taut Type 2 curves.

Defining Several Important Shapes

#

Now that the framework for which curves are worth considering has been well established, let’s build up a vocabulary of specific curves to consider. First, what exactly do I mean by “curve”? In a context where curves can be translated and rotated, it makes sense to define a curve more broadly than usual.

Which curves are we going to give names to? In some sense, the simplest unit-length curve is the single line segment. I don’t want to use the letter L or I for this purpose, so I’ll call this $G$, in reference to the French spelling of the word line, or ligne.

There’s a bit of an opposite to $G$. While $G$ has a maximum distance from end to end (or diameter), it has no width whatsoever. A correlating question would be to search for the maximum radius of a circle accomodated by a single curve’s convex hull.

proof:
Consider the circle $c$ of radius $r$ defined with $(y-r)^2+x^2=(r)^2$. Suppose $m$ is a curve with endpoints $a$ and $b$ that accomodates $c$ with minimal length. In order to minimize the length of $m$, $\langle a,b \rangle$ must be tangent to $c$. Without loss of generality, let $a$ and $b$ lie on the $x$ axis. The locations of $a$ and $b$ that minimize the length of $m$ are directly beneath the ends of $c$, at $(\pm r,0)$

To minimize the size of $m$, assume that $m$ is “taut” against the circle $c$, as shown in the drawing below:

The total length of $m$ is therefore $2r+\pi r$. If $m$ is a unit length curve, then $$\begin{align*} 1 =2r+\pi r &=r(2+\pi)\\ \frac{1}{2+\pi} &= r \end{align*}$$

Now that that’s done, we can use this as the basis of the definition of our second useful curve!

That’s the largest possible circle that can be accomodated with a unit length curve, but what about the largest possible square? We might as well define it as simply as possible:

Of course, $S$ was chosen as a name because that is the first letter of the word “square”. This shape is shown below:

What other simple curves are important enough to be given brief names? There’s a fair argument to be made that the simplest curves would be those of constant curvature, also known as arcs.

It’s worth showing that the shape given in this definition does in fact have a length of 1. The two places where $y=0$ are given with:

$$\begin{align*} 0 &= \sqrt{r^{2}-x^{2}}-r\sin\left(\frac{\pi r-1}{2r}\right)\\ r\sin\left(\frac{\pi r-1}{2r}\right) &= \sqrt{r^{2}-x^{2}}\\ r^2\sin^2\left(\frac{\pi r-1}{2r}\right) &= r^2-x^2\\ r^2 - r^2\sin^2\left(\frac{\pi r-1}{2r}\right) &= x^2\\ r^2\left(1-\sin^2\left(\frac{\pi r-1}{2r}\right)\right) &= x^2\\ r^2\cos^2\left(\frac{\pi r-1}{2r}\right) &= x^2\\ \pm r\cos\left(\frac{\pi r-1}{2r}\right) &= x \end{align*}$$

Note also that $y’=\frac{x}{\sqrt{r^{2}-x^{2}}}$. Using the standard formula for length of a curve gives:

$$\begin{align*} \int_{-r\cos\left(\frac{\pi r-1}{2r}\right)}^{r\cos\left(\frac{\pi r-1}{2r}\right)}\sqrt{1+y’^2}dx &= \int_{-r\cos\left(\frac{\pi r-1}{2r}\right)}^{r\cos\left(\frac{\pi r-1}{2r}\right)}\sqrt{1+\frac{x^2}{r^2-x^2}}dx\\ &= \int_{-r\cos\left(\frac{\pi r-1}{2r}\right)}^{r\cos\left(\frac{\pi r-1}{2r}\right)}\sqrt{\frac{r^2}{r^2-x^2}}dx\\ &= \int_{-r\cos\left(\frac{\pi r-1}{2r}\right)}^{r\cos\left(\frac{\pi r-1}{2r}\right)}\frac{1}{\sqrt{1-\frac{x^2}{r^2}}}dx\\ &= \left.-r\cos^{-1}\left(\frac{x}{r}\right)\right|_{-r\cos\left(\frac{\pi r-1}{2r}\right)}^{r\cos\left(\frac{\pi r-1}{2r}\right)}\\ &= -r\cos^{-1}\left(\frac{r\cos\left(\frac{\pi r-1}{2r}\right)}{r}\right) + r\cos^{-1}\left(\frac{-r\cos\left(\frac{\pi r-1}{2r}\right)}{r}\right)\\ &= -r\cos^{-1}\left(\cos\left(\frac{\pi r-1}{2r}\right)\right) + r\cos^{-1}\left(-\cos\left(\frac{\pi r-1}{2r}\right)\right)\\ &= -r\left(\frac{\pi r-1}{2r}\right) + r\left(-\cos^{-1}\left(\cos\left(\frac{\pi r-1}{2r}\right)\right)+\pi\right)\\ &= -r\left(\frac{\pi r-1}{2r}\right) + r\left(-\left(\frac{\pi r-1}{2r}\right)+\pi\right)\\ &= -\frac{\pi r-1}{2} -\frac{\pi r-1}{2}+\pi r\\ &= -\pi r+1 +\pi r = 1 \end{align*}$$

One could also argue that the simplest shape to consider is two line segments meeting at a single vertex, also known as a 2-chain (no relation), with both line segments of equal length. The convex hulls of these shapes are simple triangles. From this perspective, it would only make sense to define

As you might have guessed, the name $V$ was chosen due to its resemblance to an angle. The name $E$ was chosen because of the similarity to an equilateral triangle. As before, it’s worthwhile to show that the length of this curve is 1. The distance between $\left(0,\frac{1}{2}\sin{\frac{\theta}{2}}\right)$ and $\left(\frac{1}{2}\cos{\frac{\theta}{2}},0\right)$ is given by:

$$\sqrt{\left(\frac{1}{2}\sin{\frac{\theta}{2}}\right)^2 + \left(\frac{1}{2}\cos{\frac{\theta}{2}}\right)^2} = \sqrt{\frac{1}{4}\sin^2{\frac{\theta}{2}} + \frac{1}{4}\cos^2{\frac{\theta}{2}}} = \frac{1}{2}$$

All of these have been type 1 curves! In a sense, the simplest type 2 curve is one that can be formed with just four points.

Of course, this is also shown below:

A hopeful reader might wonder if this curve might in fact be accomodated by a type 1 curve. $Z$ certainly doesn’t look like it could be accomodated by a type 1 curve, by any of the lemmas encountered earlier.

To see if it’s possible to accomodate this with a type 1 curve, let’s start by placing the $Z$ curve on it’s side, so that the starting and ending points are on the $y$-axis. I will refer to this instance of $Z$ as $z$, and label its points with:

$$z_1 = \left(0,0\right)$$ $$z_2 = \left(\frac{2\sqrt{5}}{15},-\frac{\sqrt{5}}{15}\right)$$ $$z_3 = \left(\frac{3\sqrt{5}}{15},\frac{\sqrt{5}}{15}\right)$$ $$z_4 = \left(\frac{\sqrt{5}}{3},0\right)$$

I’m loath to stretch the length of this paper out any further with information the reader could easily find for themselves. That said, it is important to show that these points really do form an instance of $Z$. First, note that the distances between each point really does come out to ⅓.

$$\ell\langle z_1,z_2\rangle = \sqrt{\left(\frac{2\sqrt{5}}{15} -0\right)^2 + \left(-\frac{\sqrt{5}}{15} -0\right)^2} = \sqrt{\frac{20}{15^2} + \frac{5}{15^2}} = \frac{5}{15}=\frac{1}{3}$$

$$\ell\langle z_2,z_3\rangle = \sqrt{\left(\frac{3\sqrt{5}}{15} -\frac{2\sqrt{5}}{15}\right)^2 + \left(\frac{\sqrt{5}}{15} +\frac{\sqrt{5}}{15} \right)^2} = \sqrt{\frac{5}{15^2} + \frac{20}{15^2}} = \frac{5}{15}=\frac{1}{3}$$

$$\ell\langle z_3,z_4\rangle = \sqrt{\left(\frac{5\sqrt{5}}{15} -\frac{3\sqrt{5}}{15}\right)^2 + \left(\frac{0-\sqrt{5}}{15}\right)^2} = \sqrt{\frac{20}{15^2} + \frac{5}{15^2}} = \frac{5}{15}=\frac{1}{3}$$

Additionally, the slope between $z_1$ and $z_2$ is $\frac{-\frac{\sqrt{5}}{15}-0}{\frac{2\sqrt{5}}{15}-0} = \frac{-1}{2}$. The slope between $z_2$ and $z_3$ is $\frac{\frac{\sqrt{5}}{15}+\frac{\sqrt{5}}{15}}{\frac{3\sqrt{5}}{15}-\frac{2\sqrt{5}}{15}} = 2$. Finally, the slope between $z_4$ and $z_3$ is $\frac{0-\frac{\sqrt{5}}{15}}{\frac{\sqrt{5}}{3}-\frac{3\sqrt{5}}{15}} = \frac{-1}{2}$. This shows that each edge is orthogonal to the previous one, confirming that this is an instance of $Z$.

I want to define a type 1 curve that gets very close to accomodating this. For a start, I’ll have this curve include $z_1$, $z_2$, and $z_3$. This curve is shown in purple:

The length of $\langle z_1, z_3, z_4\rangle$ is given by:

$$\ell\langle z_1, z_3\rangle + \ell\langle z_3, z_4\rangle = \sqrt{\left(\frac{\sqrt{5}}{15}-0\right)^2+\left(\frac{3\sqrt{5}}{15}-0\right)^2} + \frac{1}{3} = \sqrt{\frac{5}{15^2}+ \frac{45}{15^2}} + \frac{5}{15} = \frac{5\sqrt{2}+5}{15}=\frac{\sqrt{2}+1}{3}$$

In order to get as close as possible to accomodating $z_2$, we may place a single point at a distance of $\frac{2-\sqrt{2}}{3}$ from $z_1$. The optimal point $p$ would be the one where $\langle z_1,p \rangle$ and $\langle p, z_4 \rangle$ are orthogonal. (trust me)

If $p = \left(p_x, p_y\right)$, then these two requirements may be expressed as:

$$\sqrt{\left(p_x-0\right)^2+\left(p_y-0\right)^2} = \frac{2-\sqrt{2}}{3}$$

$$\frac{p_y-0}{p_x-0}=-\frac{\frac{\sqrt{5}}{3}-p_x}{0-p_y}$$

If we take the first of these two equations and square both sides, this becomes:

$$p_x^2+p_y^2 = \frac{6-4\sqrt{2}}{9}$$

Meanwhile, we multiply both sides of the other equation by $p_xp_y$ to get

$$p_y^2=\frac{\sqrt{5}}{3}p_x-p_x^2$$

Now we can plug this into the other equation for $p_y^2$ and solve to get

$$\frac{\sqrt{5}}{3}p_x = \frac{6-4\sqrt{2}}{9}$$ $$p_x = \frac{6-4\sqrt{2}}{3\sqrt{5}}$$

Square this to get

$$p_x^2 = \frac{68-48\sqrt{2}}{45}$$

Plug this and $\frac{\sqrt{5}}{3}p_x$ into our equation for $p_y^2$ to get

$$\begin{align*} p_y^2 &= \frac{6-4\sqrt{2}}{9}-\frac{68-48\sqrt{2}}{45}\\ p_y^2 &= \frac{30-20\sqrt{2}}{45}-\frac{68-48\sqrt{2}}{45}\\ p_y^2 &= \frac{-38+28\sqrt{2}}{45}\\ p_y &= \pm\frac{\sqrt{-38+28\sqrt{2}}}{3\sqrt{5}} \end{align*}$$

Of course, this is pretty ugly, but it is accurate. We’ll use the negative version of $p_y$ to get $p=\left(\frac{6-4\sqrt{2}}{3\sqrt{5}},\frac{\sqrt{-38+28\sqrt{2}}}{3\sqrt{5}}\right)$, as shown:

Convex Hulls of Multiple Curves

#

My plan is to spend this section establishing some useful lemmas about the convex hulls of multiple curves, as a bit of set dressing for the sections to come. The nature of this problem requires that we seek to minimize the size of a convex hull, so I’ll give a definition relating to this:

Here’s the first lemma for this section:

proof:
Let $a’$ be the instance of $A$ that is internal to $X$ but does not touch its boundary, let $\omega’=\lbrace a’, b, c, \dots\rbrace$, and let $X’$ be the convex hull of the union of $\omega’$.

It can be immediately seen by definition that $\omega’\in\Omega$, and therefore that $X’\in [\Omega]$. Because $X$ is a minimal element of $[\Omega]$, $X\lesssim X’$. $X’$ contains no points outside the boundary of $X$, and so $X\nless X’$.

Therefore, $X’$ has the same area as $X$. This implies that $X=X’$.

In order to produce the next lemma. we give a special definition:

This definition comes with an almost immediate lemma:

proof:

Let $a,b\in\omega$. First, we know that $[\omega]\sqsupseteq[a,b]$. We also know that if $a$ and $b$ are not fully intersecting, they at least have a fully intersecting pair of linear transformations, $a’$ and $b’$, such that $[a,b]\sqsupseteq[a’,b’]$. By the transitive property, $[\omega]\sqsupseteq[a’,b’]$

We can extend this lemma just a little bit further, finally reaching a theorem about the nature of minimal convex hulls.

proof:

Let $\omega$ be such that $[\omega]\in[\Omega]^-$. In order to force a contradiction, assume that there exists $a,b\in\omega$ such that $a$ and $b$ are not fully intersecting. Because of Lemma 6, we may assume that both $a$ and $b$ touch the boundary of $[\omega]$. In fact, they both extend the boundary of $[\omega]$, meaning that removing them would result in a smaller overall convex hull.

Let’s assume (WLOG) that $a’$ is a linear transformation of $a$, such that $a’\sqsubset[a,b]$. Let $\omega’$ be formed by replacing $a$ in $\omega$ with $a’$. Note that, by the definition of fully intersecting, $[a,b]\sqsupseteq[a’b]$, and therefore, $[\omega]\sqsupseteq[\omega’]$. This means that if there is any point on the boundary of $[\omega]$ that is not present in $[\omega’]$, then $[\omega]\sqsupset[\omega’]$. But this cannot be the case if $[\omega]\in[\Omega]^-$, so it must be that $[\omega]=[\omega’]$.

We now know that there are some points present in $[a,b]$ that are absent in $[a’,b]$, but that none of these points are on the boundary of $[\omega]$. As mentioned, $a$ lies on the boundary of $[\omega]$. If $a’$ is also on the boundary of $[\omega]$, that means that $a$ must not have actually expanded the boundary of $[\omega]$. This also violates the spirit of Lemma 6, giving a contradiction.

For this lemma to be maximally useful, it’s a good idea to characterize better what makes two curves fully intersecting.

proof:

To use more down-to-earth terminology, if $[c,d]/\left(c\cup d\right)$ is discontinuous, this means that $[c,d]$ adds two non-adjacent regions to $\left(c\cup d\right)$. An example of this is shown below. The dark grey regions are the interiors of $c$ and $d$, while the light grey regions are the areas added by $[c,d]$

Let $\left\langle c_1,d_1\right\rangle$ and $\left\langle c_2,d_2\right\rangle$ be the two lines on the border of $[c,d]$ that are not part of $c$ or $d$, where $c_1$ and $c_2$ lie on $c$, and $d_1$ and $d_2$ lie on $d$. Assume, without loss of generality, that $\ell\left\langle c_1,c_2\right\rangle \leq \ell\left\langle d_1,d_2\right\rangle$, so that the two lines are either parallel, or slanting to be closer near $c$. This is shown below:

For what comes next, consider rotating $[c,d]$ such that the slope of $\left\langle c_1,d_1\right\rangle$ is non-negative, and the slope of $\left\langle c_2,d_2\right\rangle$ is non-positive, as in the previous diagram. For every point $p$ in $c$, there is a positive distance between $p$ and the point on the border of $d$ and $[c,d]$ that shares $p$’s $y$-coordinate. If there wasn’t, then $p$ would lie on the the border of $d$, violating one of our assumptions. Let $x$ be the minimal possible such distance, and let $c’$ be the result of shifting $c$ a distance of $x$ towards $d$.

We have now demonstrated that $[c’,d]\subset[c,d]$, and by definition, $c$ and $d$ are not fully intersecting.

proof:

Let $g_1$ be the endpoint of $g$ that is internal to $c$, meaning that it is in $c$, but not on its boundary. Let $g_2$ be the endpoint of $g$ that is external to $c$. There exists a closed ball with positive radius $r$ around $g_1$ entirely in $c$. Let $g_2’$ be the point on $g$ at distance $r$ from $g_2$, and let $g_1’$ be the point 1 unit from $g_2’$ such that $\left\langle g_1’,g_2’\right\rangle$ passes through $g_1$. furthermore, let $g’=\left\langle g_1’,g_2’\right\rangle$. We may observe that $g_1’$ an $g_2’$ are both in $[c,g]$, but $g_2$ is not in $[c,g’]$. By definition, $[c,g’]\subset[c,g]$, and so $g$ and $c$ are not fully intersecting.

For the final result of this this section, it’ll be helpful to note one final definition. To be clear, this isn’t a definition of my own, but one that is commonly used.

The theorem we’re attempting to develop with this is:

Showing this is quite hard! It seems true, but that’s hardly proof. I’m spinning off this discussion into its own mini-article, so it doesn’t become a massive part of this one.

Moserville

16 April 2025·558 words·3 mins

Examples of Minimal Covers on Finite Sets

#

In a previous section, I limited the space of curves worth considering by defining taughtness. I showed that some curves are accomodated by other curves, and therefore not worth considering. In the next section, I’ll try to limit the space of curves even further by finding some curves that are accomodated by sets of other curves.

My initial goal was to hopefully find a finite set of taut curves that proxy accommodate all other taut curves, and then to find the smallest set that contains all curves in that finite set. This goal seems reasonable at first, as it is much easier to find $[\Xi]^-$ when $\Xi$ is a finite set of curves as opposed to an infinite one. There are some severe problems with this as an objective, but there’s still much to be learned by exploring it!

In this section, to illustrate that it is in fact possible to find the minimal size of covering shapes for certain finite numbers of curves, we’ll find a few examples. These next two theorems are 5% insight and 95% bookkeeping, so feel free to skip their proofs if you’re in a hurry.

proof:
First, assume without loss of generality that the instance of $S$ is located on the points $(0,0)$, $\left(\frac{1}{3},0\right)$, $\left(\frac{1}{3},\frac{1}{3}\right)$, and $\left(0,\frac{1}{3}\right)$, as in the definition’s example. Call this instance $s$. Let the intance of $G$ have endpoints $A$ and $B$, so that $\ell\langle A,B\rangle = 1$. Define this instance of the line so that $\langle A,B\rangle = g$. Note that the line between $A$ and $B$ must pass through the square in order for these shapes to be fully intersecting.

Of course, $[s,g]$ is the convex hull of these points. A potential instance of $[s,g]$ is shown in the illustration below:

$[s,g]$ can be deconstructed into the central square and multiple triangles:

Let $A=(a_x,a_y)$, and let $B=(b_x,b_y)$. Additionally, let $f(x)$ be the distance of $x$ to the interval $\left[0,\frac{1}{3}\right]$, so that

$$f(x) = \begin{cases} -x &\text{if } x\leq 0 \\ 0 &\text{if } x\in \left(0,\frac{1}{3}\right)\\ x-\frac{1}{3} &\text{if } x \geq \frac{1}{3} \end{cases}$$

To give an example of how this could be useful, consider the triangle formed by $a_x$. If $A$ is the lower point on the figure above, this would be the green triangle. Using the standard formula for area of a triangle, we get $\frac{1}{2}\cdot\frac{1}{3}\cdot f(a_x)$

Then, the full area of $[s,g]$ would be given by:

$$\frac{1}{9}+\sum_{i} \frac{f(i)}{6}$$

Where $i$ is an element of $\left\lbrace a_x,a_y,b_x,b_y\right\rbrace$. The presence of $\frac{1}{9}$ is due to the area of $s$. There is also the limitation from the length of $g$ which implies that:

$$\begin{align*} 1 &= \ell \langle A,B\rangle\\ 1 &= \sqrt{\left(a_x-b_x\right)^2+\left(a_y-b_y\right)^2}\\ 1 &= \left(a_x-b_x\right)^2+\left(a_y-b_y\right)^2 \end{align*}$$

For the time being, let’s hold $A$ constant, and consider the optimal placement of $B$, assuming without loss of generality that $b_x \geq 0$ and that $b_y\geq \frac{1}{3}$. Solving the previous equation for $b_y$ gives

$$\begin{align*} 1 &= \left(a_y-b_y\right)^2+\left(a_x-b_x\right)^2\\ 1- \left(a_y-b_y\right)^2 &= \left(a_x-b_x\right)^2\\ \sqrt{1- \left(a_y-b_y\right)^2} &= a_x-b_x\\ a_x - \sqrt{1- \left(a_y-b_y\right)^2} &= b_x \end{align*}$$

Therefore, the maximum value of $b_x$ is:

$$b_x \leq a_x - \sqrt{1- \left(a_y-\frac{1}{3}\right)^2}$$

By symmetry, we also know that $b_y= a_y - \sqrt{1- \left(a_x-b_x\right)^2}$. Therefore, the total area due to $B$ can be expressed as:

$$A_B = f(b_x) + f(b_y) = f(b_x) + b_y = f(b_x) + a_y - \sqrt{1- \left(a_x-b_x\right)^2}$$

This function is non-differentiable or has boundaries when $b_x=0$, when $b_x=\frac{1}{3}$, and when $b_x= a_x - \sqrt{1- \left(a_y-\frac{1}{3}\right)^2}$. If $x\in\left[0,\frac{1}{3}\right]$, then

$$A_B = a_y - \sqrt{1- \left(a_x-b_x\right)^2}$$

The derivative of this is

$$\frac{dA_B}{db_x} = \frac{2a_x-2b_x}{2\sqrt{1- \left(a_x-b_x\right)^2}} = \frac{a_x-b_x}{\sqrt{1- \left(a_x-b_x\right)^2}}$$

This is equal to 0 only when $b_x = a_x$. On the other hand, if $x > \frac{1}{3}$, then

$$A_B = b_x-\frac{1}{3} + a_y - \sqrt{1- \left(a_x-b_x\right)^2}$$

The derivative of this is quite similar to the derivative from before:

$$\frac{dA_B}{db_x} = 1 + \frac{a_x-b_x}{\sqrt{1- \left(a_x-b_x\right)^2}}$$

For this derivative to be equal to 0, this comes to:

$$\begin{align*} 0 &= 1 + \frac{a_x-b_x}{\sqrt{1- \left(a_x-b_x\right)^2}}\\ 1 &= \frac{b_x-a_x}{\sqrt{1- \left(a_x-b_x\right)^2}}\\ \sqrt{1- \left(a_x-b_x\right)^2} &= b_x-a_x\\ 1- \left(a_x-b_x\right)^2 &= \left(b_x-a_x\right)^2\\ 1 &= 0 \end{align*}$$

This is obviously not possible, meaning that the critical points of this area function are when $b_x=0$, when $b_x=\frac{1}{3}$, when $b_y=\frac{1}{3}$, or when $0 < b_x=a_x < 0$.

If $b_x,a_x \in \left[0,\frac{1}{3}\right]$, then the formula for area is given by $a_y - \sqrt{1- \left(a_x-b_x\right)^2}$. It is clear from inspection that this function is larger when $a_x=b_x$ as opposed to otherwise. Because of this, we conclude that $B$ lies on either $x=0$, $x=\frac{1}{3}$, $y=0$, or $y=\frac{1}{3}$. By symmetry, $A$ also lies on one of these lines. This leads to one of two possibilities:

In one possibility, assume that both $A$ and $B$ lie on $x=0$ and $y=0$. Without loss of generality, let $b_x = 0$ while $a_x = \frac{1}{3}$. Then the distance between \(A) and \(B\) gives the restriction that:

$$\begin{align*} 1 &= \sqrt{\left(a_x-b_x\right)^2+\left(a_y-b_y\right)^2}\\ 1 &= \sqrt{\left(\frac{1}{3}\right)^2+\left(a_y-b_y\right)^2}\\ 1 &= \frac{1}{9}+\left(a_y-b_y\right)^2\\ \frac{8}{9} &= \left(a_y-b_y\right)^2\\ \frac{2\sqrt{2}}{3} &= |a_y-b_y| \end{align*}$$

Assuming that the square lies between these two points, the area of the full shape is then given by:

$$A= \frac{1}{9}+\frac{1}{6}\cdot\frac{2\sqrt{2}}{3}=\cdot\frac{2+2\sqrt{2}}{18}\approx 0.268$$

In the other possibility, $A$ lies on a vertical line, while $B$ lies on a horizontal line, or vice versa. Without loss of generality, let $a_x=0$, and let $b_y = 0$.Then the distance between \(A) and \(B\) gives the restriction that:

$$\begin{align*} 1 &= \sqrt{\left(a_x-b_x\right)^2+\left(a_y-b_y\right)^2}\\ 1 &= \sqrt{b_x^2+a_y^2}\\ 1 &= b_x^2+a_y^2\\ 1 - a_y^2 &= b_x^2\\ \sqrt{1 - a_y^2} &= b_x \end{align*}$$

And the total area of the shape is given by:

$$A= \frac{1}{9}+\frac{1}{6}b_x + \frac{1}{6}a_y=\frac{1}{9}+\frac{\sqrt{1 - a_y^2}}{6} + \frac{a_y}{6}$$

Once again, we find the minimum of this using first-term calculus. The derivative of this area is:

$$\frac{dA}{da_y}= \frac{ -a_y}{6\sqrt{1 - a_y^2}}+\frac{1}{6}$$

This derivative evaluates to 0 when

$$\begin{align*} 0 &= \frac{ -a_y}{6\sqrt{1 - a_y^2}}+\frac{1}{6}\\ \frac{a_y}{6\sqrt{1 - a_y^2}} &= \frac{1}{6}\\ a_y &= \sqrt{1 - a_y^2}\\ a_y^2 &= 1 - a_y^2\\ 2a_y^2 &= 1\\ a_y &= \frac{1}{\sqrt{2}} \end{align*}$$

If $a_y=\frac{1}{\sqrt{2}}$, then $b_x=\frac{1}{\sqrt{2}}$ as well. This solution is invalid, however, as the instance of $L$ in this case would not pass through the instance of $S$. Rather than showing this mathematically, I’ll just show the results of a graphing utility.

The non-differentiable points in the second possibility are identical to the non-differentiable points in the first possibility, proving the theorem.

Admittedly, the statement of that theorem might have been highly intuitive to you. You might find it much less intuitive to consider elements of $\left[G,V_{\frac{\pi}{6}}\right]^-$.

proof:

To start, let’s place an instance of $E$ in the cartesian plane, and we’ll call this instance $e$ (Euler’s number does not appear anywhere in this article). I’ll place this triangle with two points on the $x$-axis, and one point on the $y$-axis. Then the points in this triangle are:

$$e_1=\left(\frac{-1}{4},0\right)$$ $$e_2=\left(\frac{1}{4}\right)$$ $$e_3=\left(0,\frac{\sqrt{3}}{4}\right)$$

Note that it isn’t really important which two sides of this triangle make up $e$, as we’ll be considering the convex hull, so all three sides are included in this graph. I will illustrate this triangle below:

Note that I have labeled the sections of this triangle with numbers 1 though 6. This is because the instance of $G$ must pass through $e$ by Theorem 4. Note that $G$ cannot have endpoints in adjacent sections if it passes through $e$. Note also that $G$ cannot have both its endpoints in odd-numbered sections. Therefore, define the endpoints of $G$ to be $a=\left(a_x,a_y\right)$ and $b=\left(b_x,b_y\right)$. Without loss of generality, assume $a$ is in section 4, and assume $b$ is in section 1 or 6.

What is the size of $\left[e,\langle a,b\rangle\right]$? First, Let’s find the area of $\left[e,a\right]$. This consists of the area inside $e$, as well as the area in the triangle defined by the points $e_1$, $e_2$, and $a$.

The area of the grey region inside $e$ is given by $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{\sqrt{3}}{4}$. The area of the blue region created by $a$ is given by $\frac{1}{2}\cdot\frac{1}{2}\cdot a_y$. Therefore, the total area of these two regions is:

$$\frac{\sqrt{3}}{16}+\frac{a_y}{4}$$

Let’s start by assuming that $b$ is located in region 6. In this case, the inclusion of $b$ creates an additional triangle defined by the points $e_3$, $e_2$, and $b$, as shown below:

The area of this green region is determined by the distance from $b$ to the line from $e_3$ to $e_2$. There’s a useful formula that gives the area of a triangle given the locations of its three points:

$$\begin{align*} Area &=\left|{\frac{b_x\left(\frac{\sqrt{3}}{4}-0\right)+0\left(0-b_y\right)+\frac{1}{4}\left(b_y-\frac{\sqrt{3}}{4}\right)}{2}}\right|\\ &= \left|{\frac{\frac{b_x\sqrt{3}}{4}+\frac{1}{4}\left(b_y-\frac{\sqrt{3}}{4}\right)}{2}}\right|\\ &= \left|{\frac{b_x\sqrt{3}+b_y-\frac{\sqrt{3}}{4}}{8}}\right|\\ &= \frac{b_x\sqrt{3}}{8}+\frac{b_y}{8}-\frac{\sqrt{3}}{32} \end{align*}$$

Recall that $b$ and $a$ are exactly one unit apart. This gives the restriction that:

$$\begin{align*} 1 &= \left(b_y-a_y\right)^2 +\left(b_x-a_x\right)^2\\ 1-\left(b_x-a_x\right)^2 &= \left(b_y-a_y\right)^2 \\ \sqrt{1-\left(b_x-a_x\right)^2} &= b_y-a_y \\ \sqrt{1-\left(b_x-a_x\right)^2}+a_y &= b_y\\ \end{align*}$$

Substituting this into the equation of the area of the green region would give

$$Area = \frac{b_x\sqrt{3}}{8}+\frac{\sqrt{1-\left(b_x-a_x\right)^2}+a_y}{8}-\frac{\sqrt{3}}{32}$$

To find the minimum possible value of this area, we take the derivative with respect to $b_x$ and set that derivative equal to 0.

$$\begin{align*} 0&= \frac{\sqrt{3}}{8}+\frac{-2\left(b_x-a_x\right)}{8\cdot 2\sqrt{1-\left(b_x-a_x\right)^2}}\\ 0&= \frac{\sqrt{3}}{8}-\frac{b_x-a_x}{8\sqrt{1-\left(b_x-a_x\right)^2}}\\ 0&= \sqrt{3}-\frac{b_x-a_x}{\sqrt{1-\left(b_x-a_x\right)^2}}\\ \sqrt{3} &= \frac{b_x-a_x}{\sqrt{1-\left(b_x-a_x\right)^2}}\\ \sqrt{3-3\left(b_x-a_x\right)^2} &= b_x-a_x\\ 3-3\left(b_x-a_x\right)^2 &= \left(b_x-a_x\right)^2\\ 3 &= 4\left(b_x-a_x\right)^2\\ \pm\frac{\sqrt{3}}{2} &= b_x-a_x\\ \end{align*}$$

Again, the distance between $b$ and $a$ must be exactly one unit. This means that:

$$\begin{align*} 1 &= \left(b_y-a_y\right)^2 +\left(\pm\frac{\sqrt{3}}{2}\right)^2\\ 1 &= \left(b_y-a_y\right)^2 + \frac{3}{4}\\ \frac{1}{4} &= \left(b_y-a_y\right)^2\\ \frac{1}{2} &= b_y-a_y\\ \end{align*}$$

Because $b_x>0$, and $b_y$ is so low above $a_y$, we may assume that $b= \left(a_x+\frac{\sqrt{3}}{2},a_y+\frac{1}{2}\right)$. This is signifigant, because we may observe that the slope from $a$ to $b$ is $\frac{1}{\sqrt{3}}$. This is perpendicular to the slope from $Z_3$ to $Z_2$, meaning that this location of $b$ maximizes the area of the green triangle.

The other posibility that would create a critical point for the area of the green triangle is that $b$ lies on either the $x$-axis, or on the line $y=\sqrt{3}x+\frac{\sqrt{3}}{4}$, as these are the bounds of $b$’s location. It is not possible for $b$ to lie on the $x$-axis, as this would result in the line $G$ lying outside $e$. Therefore, the minimum area is achieved when $b$ lies on $y=\sqrt{3}x+\frac{\sqrt{3}}{4}$.

The other posibility is that $b$ lies in section 1. If this is the case, we may ignore the area of $e$, as the overall area can now be defined by a triangle between $b$, $e_1$, and $e_2$.

In this case, the total area of $[e,\langle a,b\rangle]$ is $\frac{1}{4}a_y+\frac{1}{4}b_y$. Holding the position of $a$ constant, $b_y$ is minimized by maximizing $\left|\left(b_x-a_x\right)\right|$. This means that $b$ must lie on the boundary of Section 1.

We have concluded that $b$ must lie on the boundary of section 1, regardless of what other assumptions have been made. Let’s assume, without loss of generality, that $b$ lies on the line $y=\sqrt{3}x+\frac{\sqrt{3}}{4}$, while $a$ is in section 4. From here, we’ll hold the location of $b$ constant and find the optimal location of $a$. As before, to minimize the area of the blue triangle formed by $a$, we must minimize $a_y$, which means maximizing $\left|\left(b_x-a_x\right)\right|$. We conclude that $a$ must lie on the boundary of section 1. If $a$ lies on the line passing through $e_3$ and $e_2$, then $\langle a,b\rangle$ does not intersect $e$.

We have now shown that $[e, \langle a,b\rangle]$ is congruent to a triangle with base length 1 and height $\frac{\sqrt{3}}{4}$. The area of such a triangle is $\frac{\sqrt{3}}{4}$.

Some Difficulties in Finding Minimal Covers

#

There are two ideas that I imediately had when considering how to find minimal covers. Neither of them are effective, but it’s informative to see how these strategies can fail. The first would look like this:

Another way of phrasing this is that there might not be a way to smoothly slide a shape into position to minimize the size of the complex hull without first increasing the size of the complex hull. A third way of phrasing it is that there may be local minima in the size of the convex hull of two shapes that is not an absolute minimum.

proof

Let $b$ be a rectangle defined by the points $0,0$, $0,\frac{1}{4}$, $\frac{1}{2},\frac{1}{4}$, and $\frac{1}{2},0$. Consider the convex hull of this rectangle and a unit length line $A$. It was shown in Theorem 5 that the optimal locations for this line are such that its endpoints lie on opposite parallel extensions of the sides of $b$. This means that the endpoints of $a$ either lie on $x=0$ and $x=\frac{1}{2}$, or they lie on $y=0$ and $y=\frac{1}{4}$. Without loss of generality, assume one endpoint of $a$ lies at the origin. Because the other endpoint of $a$ must lie a unit away from the origin, the two possible locations are $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ and $\left(\frac{\sqrt{15}}{4},\frac{1}{4}\right)$, as shown below:

The area of the box on its own is $\frac{1}{2}\cdot\frac{1}{4} = \frac{1}{8}$. The total area of $[a,b]$ is:

$$\begin{align*} \frac{1}{8} + \frac{1}{2}\cdot\frac{1}{2}\cdot\left(\frac{\sqrt{3}}{2} - \frac{1}{4}\right) = \frac{2}{16} + \frac{1}{4}\cdot\frac{2\sqrt{3}-1}{4} = \frac{2\sqrt{3}+1}{16}\approx 0.279 & \text{ if an endpoint lies at } \left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) \\ \frac{1}{8} + \frac{1}{2}\cdot\frac{1}{4}\cdot\left(\frac{\sqrt{15}}{4} - \frac{1}{2}\right) = \frac{4}{32} + \frac{1}{8}\cdot\frac{\sqrt{15}-2}{4} = \frac{\sqrt{15}+2}{32}\approx 0.184 & \text{ if an endpoint lies at } \left(\frac{\sqrt{15}}{4},\frac{1}{4}\right) \\ \end{align*}$$

Suppose $a$ is located with an endpoint at $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$. A transformation that brings $a$ to $\left(\frac{\sqrt{15}}{4},\frac{1}{4}\right)$ would have to first bring it through the area above $y=\frac{1}{4}$ and to the right of $x=\frac{1}{2}$, which would increase the size of $[a,b]$.

What a shame! I was hoping that it would be possible to smoothly shift shapes to find their best convex hulls. An even bigger issue came when I realized this:

In other words, it doesn’t work to find the minimal shell of two shapes and then find the minimal shell of that with a third shape. You have to consider all three shapes at the same time.

proof

Consider the set of curves consisting of the unit length line segment, the equilateral triangle with sides of length $\frac{1}{2}$, and the square with sides of length $\frac{1}{3}$, or ${G, E, S}$. To cause a contradiction, let $[s,g]$ be an instance of $[S,G]^-$ and let $[g’,e]$ be an instance of $[G, E]^-$, so that

$$[s,g,g’,e]\in[S, G, E]^-$$

Immediately, we have that $[s,g,e]$ is not larger than $[s,g,g’,e]$. Because $[s,g,g’,e]\in[S, G, E]^-$, we have that $[s,g,e]$ cannot be smaller than $[s,g,g’,e]$, and so they must be the same size. By the same reasoning, $[s,g’,e]$ must be the same size as $[s,g,g’,e]$ as well. This means that $g$ or $g’$ can be removed from $[s,g,g’,e]$ without shrinking it, but not both. Therefore, $g$ and $g’$ must share at least one endpoint on the boundary of $[s,g,g’,e]$. Without loss of generality, let this endpoint be the origin.

First, let’s assume that $g = g’$. Without loss of generality, we can let $g$ lie on the $x$-axis, with $e$ lying above the $x$ axis. We know that $e$ has a height of $\frac{\sqrt{3}}{4}$, and a base length of $\frac{1}{2}$, meaning that $[g,e]$ consists of a triangle between the points $(0,0)$, $(1,0)$, and $(a,\frac{\sqrt{3}}{4})$, with $a\in[\frac{1}{4},\frac{3}{4}]$.

Without loss of generality, we’ll also assume that the location of $s$ is such that the lines passing through the endpoints of $g$ have positive slope. Then the sides of $$ are defined by the equations:

$$y=\frac{\sqrt{2}}{4}x$$ $$y=\frac{\sqrt{2}}{4}(-1)$$ $$y=-2\sqrt{2}x+b$$ $$y=-2\sqrt{2}x+1+b$$

Where $b\in[0,2\sqrt{2}-1]$. This is shown below:

The bottom point of this square is located at the intersection of $y=-2\sqrt{2}x+b$ and $y=\frac{1}{2\sqrt{2}}\left(x-1\right)$. This point is located at:

$$\left(\frac{2\sqrt{2}b+1}{9}, \frac{b-2\sqrt{2}}{9}\right)$$

This point forms a triangle with the instance of $g$ on the x axis. That triangle has a size of

$$\frac{b-2\sqrt{2}}{18}$$

There is also a triangle formed by the top left point of $s$, whenever this point is not inside $[g,e]$. This point is located at

$$\left(\frac{\sqrt{8}\left(1+b\right)}{9},\frac{1+b}{9}\right)$$

This triangle only exists when the top right point of $s$ is above the top left line of $[e,g]$. This line is determined by

$$y=\frac{\sqrt{3}}{4a-4}\left(x-1\right)$$

Therefore, if this triangle exists, we may assume $$\begin{align*} \frac{1+b}{9} &\geq \frac{\sqrt{3}}{4a-4}\left(\frac{2\sqrt{2}\left(1+b\right)}{9}-1\right)\\ \left(1+b\right)\left(4a-4\right) &\geq\left(2\sqrt{2}\left(1+b\right)-9\right)\sqrt{3}\\ \end{align*}$$

If this is satisfied, there is a triangle with the point of $g$ located at $\left(1,0\right)$, and the point of $e$ located at $\left(a, \frac{\sqrt{3}}{4}\right)$ with $a\in\left[\frac{1}{4},\frac{3}{4}\right]$. Using the shoelace formula, the area of this triangle is given by

$$\begin{align*} &\frac{1}{2}\left|\left(1-\frac{2\sqrt{2}\left(1+b\right)}{9}\right)\left(\frac{\sqrt{3}}{4}-0\right)-\left(1-a\right)\left(\frac{1+b}{9}-0\right)\right|\\ &\frac{1}{72}\left|9\sqrt{3}-2\sqrt{6}\left(1+b\right)+4a-4+4ab-4b\right|\\ \end{align*}$$

Because we are assuming that this triangle exists, this may be rewritten as

$$\frac{1}{72}\left(\left(2\sqrt{2}\left(1+b\right)-9\right)\left(\sqrt{3}\right)-\left(4a-4\right)\left(1+b\right)\right)$$

Therefore, it is ideal to maximize $a$, so let $a=\frac{3}{4}$. The equation for area then simplifies to

$$\frac{1}{72}\left(\left(2\sqrt{6}+1\right)\left(1+b\right)-9\sqrt{3}\right)$$

This is a linear equation in which area grows with $b$ proportionally to \(\frac{2\sqrt{6}+1}{72}\aprox 0.0819\). This is quicker growth than the growth of the bottom triangle, which grows proportianally to \(\frac{1}{18}\aprox 0.055\). We may conclude that $b$ must be such that

$$\left(1+b\right)\left(4a-4\right) =\left(2\sqrt{2}\left(1+b\right)-9\right)\sqrt{3}$$

$$\begin{align*} \left(1+b\right)\left(4a-4\right) &=\left(2\sqrt{2}\left(1+b\right)-9\right)\sqrt{3}\\ \left(1+b\right) &= 9\sqrt{3}-2\sqrt{6}\left(1+b\right)\\ b &=\frac{9\sqrt{3}}{2\sqrt{6}+1}-1\\ \end{align*}$$

This is shown below:

It is evident that $s$ can be translated linearly along the line $y=\frac{\sqrt{3}}{4a-4}\left(x-1\right)$ to decrease the overall size of $[e,g,s]$, meaning that $[e,g,s]\not\in[E,G,S]^-$.

So suppose instead that $g\neq g’$, and that

In a certain sense, this is absolutely disasterous.

Proxy Accommodated Curves

#

As mentioned, one of the big hopes in this project is that we should be able to ignore certain curves because they are accomodated by the unions of other curves. For example, let’s look at what else might be covered by one of the elements of $[E,G]^-$.

proof

Let $t$ be the triangle described in the theorem statement.

For any angle $\theta$, define $v_\theta$ to contain the points $v_1=\left(0,\frac{1}{2}\cos{\frac{\theta}{2}}\right)$, $v_2=\left(\frac{1}{2}\sin{\frac{\theta}{2}},0\right)$, and $v_3=\left(-\frac{1}{2}\sin{\frac{\theta}{2}},0\right)$. Note first that the distance between $v_1$ and $v_2$ is:

$$\sqrt{\left(\frac{1}{2}\cos{\frac{\theta}{2}}\right)^2+\left(-\frac{1}{2}\sin{\frac{\theta}{2}}\right)^2}=\frac{1}{2}\sqrt{\cos^2{\frac{\theta}{2}}+\sin^2{\frac{\theta}{2}}}=1$$

By symmetry, the distance from The angle formed by this shape is $\frac{1}{2}$. The angle formed at $v_1$ is $\theta$, confirming that this is an instance of $V_\theta$. Note that $\frac{1}{2}\sin{\frac{\theta}{2}} < \frac{1}{2}$ for all values of $\theta$, meaning that $v_2$ lies inside $t$. By symmetry, $v_2$ also lies inside $t$.

$v_1$ lies inside $t$ whenever $\frac{1}{2}\cos{\frac{\theta}{2}}<\frac{\sqrt{3}}{4}$. This is true when $\theta<\frac{pi}{6}$. Since $V_\theta$ is not taut for $\theta < \frac{\pi}{6}$, we won’t consider these anyway.

For any radius $r$, let $a_r$ be the curve defined with $y=\sqrt{r^{2}-x^{2}}-r\sin\left(\frac{\pi r-1}{2r}\right)$, where $y>0$. The line that defines the right half of $t$ is $y=-\frac{\sqrt{3}}{2}x+\frac{\sqrt{3}}{4}$. These two lines intersect when

$$\begin{align*} \sqrt{r^{2}-x^{2}}-r\sin\left(\frac{\pi r-1}{2r}\right) &= -\frac{\sqrt{3}}{2}x+\frac{\sqrt{3}}{4}\\ \sqrt{r^{2}-x^{2}}-r\sin\left(\frac{\pi r-1}{2r}\right) &= \frac{\sqrt{3}}{4}(1-2x)\\ \sqrt{r^{2}-x^{2}} &= \frac{\sqrt{3}}{4}(1-2x)+r\sin\left(\frac{\pi r-1}{2r}\right)\\ r^{2}-x^{2} &= \frac{3}{16}(1-2x)^2+\frac{\sqrt{3}}{2}(1-2x)r\sin\left(\frac{\pi r-1}{2r}\right)+r^2\sin^2\left(\frac{\pi r-1}{2r}\right)\\ r^{2}-x^{2} &= \frac{3}{16}-\frac{3}{8}x+\frac{3}{4}x^2+\frac{\sqrt{3}r}{2}(1-2x)\sin\left(\frac{\pi r-1}{2r}\right)+r^2\sin^2\left(\frac{\pi r-1}{2r}\right)\\ r^2 &= \frac{3}{16}-\frac{3}{8}x+\frac{7}{4}x^2+\frac{\sqrt{3}r}{2}(1-2x)\sin\left(\frac{\pi r-1}{2r}\right)+r^2\frac{1-\cos\left(\frac{\pi r-1}{r}\right)}{2}\\ 0 &= \frac{3}{8}-\frac{3}{4}x+\frac{7}{2}x^2+\sqrt{3}r(1-2x)\sin\left(\frac{\pi r-1}{2r}\right)-r^2-r^2\cos\left(\frac{\pi r-1}{r}\right)\\ \end{align*}$$

Eliminating Type 2 Curves (hopefully)

#

So far, the previous sections have each had to do with results I have already proven. At the time that I write this, the subject of this section is not yet proven. Therefore, what follows is a hypothesis, rather than a lemma, theorem, or statement.

This would be a lovely result to find, although it’s easier stated than proven. Let’s start with a simpler hypothesis. Maybe we can use that as a stepping stone.

And in particular, I would sharpen this hypothesis even further:

To start exploring this, I’ll place an instance of $Z’$ within the cartesian plane at the same location as in its definition, called $z’$. As a reminder, the points used here are labeled as follows:

$$P = \left(p_x,p_y\right)=\left(\frac{6-4\sqrt{2}}{3\sqrt{5}},\frac{\sqrt{-38+28\sqrt{2}}}{3\sqrt{5}}\right)$$ $$Z_1=\left(0,0\right)$$ $$Z_3=\left(\frac{3\sqrt{5}}{15},\frac{\sqrt{5}}{15}\right)$$ $$Z_4=\left(\frac{\sqrt{5}}{3},0\right)$$

For convenience, we also recall that

$$Z_3=\left(\frac{2\sqrt{5}}{15},\frac{-\sqrt{5}}{15}\right)$$

It is possible, for the purposes of exploring Question 2.2, to expend the effective area of $z’$

If this shape is part of a union that includes certain other points, it would be possible for $Z$ to be included, answering Question 2.2. I’m going to indicate in blue the points which would result in accomodating $Z$. First, it should be clear that including $\left(\frac{2\sqrt{5}}{15},\frac{-\sqrt{5}}{15}\right)$would allow for $Z$. In addition, there’s an infinite stretch of territory that would include that point by necessity.

It is important to give the precise equations for these lines, as they’ll be built upon moving forward. One boundary of the blue region is the line from $P$ to $Z_2$. This is defined with:

$$y_1\leq\frac{15p_y+\sqrt{5}}{15p_x-2\sqrt{5}}\left(x-p_x\right)+p_y$$

The other boundary of the blue region is the line from $Z_2$ to $Z_4$

$$y_2\leq\frac{1}{3}\left(x-\frac{2\sqrt{5}}{15}\right)-\frac{\sqrt{5}}{15}$$

Recall that $M$ includes a circle of radius $\frac{1}{\pi+2}$. Because of this, the center of that circle cannot be located anywhere less than $\frac{1}{\pi+2}$ units away from this blue region. So, what is the line consisting of points $\frac{1}{\pi+2}$ away from $y_1$? The vector with a slope that is the opposite recipricoal of the slope of $y_1$ is:

$$v_1=\left[15p_y+\sqrt{5},-15p_x+2\sqrt{5}\right]$$

The magnitude of this vector is:

$$\begin{align*} |v_1| &= \sqrt{\left(15p_y+\sqrt{5}\right)^2+\left(-15p_x+2\sqrt{5}\right)^2}\\ &= \sqrt{225p_y^2+30\sqrt{5}p_y+5+225p_x^2-60\sqrt{5}p_x+5}\\ &= \sqrt{225p_y^2+30\sqrt{5}p_y+225p_x^2-60\sqrt{5}p_x+10} \end{align*}$$

The line that lies $\frac{1}{\pi+2}$ units away from $y_1$ is therefore:

$$y\leq\frac{15p_y+\sqrt{5}}{15p_x-2\sqrt{5}}\left(x-p_x-\frac{15p_y+\sqrt{5}}{(\pi+2)|v_1|}\right)+p_y-\frac{15p_x-2\sqrt{5}}{(\pi+2)|v_1|}$$

Similarly, if we define

$$v_2=\left[-1,3\right]$$

Then the line that lies $\frac{1}{\pi+2}$ units away from $y_1$ is:

$$y\leq\frac{1}{3}\left(x-\frac{2\sqrt{5}}{15}+\frac{1}{(\pi+2)\sqrt{10}}\right) -\frac{\sqrt{5}}{15}+ \frac{3}{(\pi+2)\sqrt{10}}$$

There will also be a circle of radius $\frac{1}{\pi+2}$ around $Z_2$. I’ll illustrate this “forbidden zone” in black. Let $f_1$ be the name of the boundary of this black region.

Next, I’ll notice that Wherever we place this instance of $M$, it must be fully intersecting with $z’$. If $Z_4$ is in the interior of $M$, then $M$ is not fully intersecting with $z’$. Similarly, if both $Z_1$ and $P$ are in the interior of $M$, $M$ is not fully intersecting with $z’$. I’ll illustrate these regions in black as well.

What is the line with the greatest slope passing through both $Z_1$ and $M$? There’s a whole host of ways to solve this question, but I’ve found that this is the quickest one. Let’s define $p,q$ to be center of the circle inside $M$, so that our diagram must include all points in the circle defined by

$$(y-q)^2+(x-p)^2=\frac{1}{\left(2+\pi\right)^2}$$

Any line passing through $Z_1$ has the form $y=mx$ where $m$ is the slope. At what locations does this pass through the circle? Substituting this into the previous equation gives:

$$\begin{align*} (mx-q)^2+(x-p)^2 &=\frac{1}{\left(2+\pi\right)^2}\\ m^2x^2-2qmx+q^2+x^2-2px+p^2 &=\frac{1}{\left(2+\pi\right)^2}\\ (m^2+1)x^2+(-2qm-2p)x+q^2+p^2 -\frac{1}{\left(2+\pi\right)^2}&=0 \end{align*}$$

The line is tangent to the circle when this equation has only one solution. Using the discriminant gives:

$$\begin{align*} (-2qm-2p)^2-4(m^2+1)\left(q^2+p^2 -\frac{1}{\left(2+\pi\right)^2}\right) &= 0\\ 4q^2m^2+8qpm+4p^2-4(m^2+1)\left(\frac{1}{\left(2+\pi\right)^2}-q^2-p^2\right) &= 0\\ q^{2}m^{2}+2qpm+p^{2}-(m^2+1)\left(q^2+p^2-\frac{1}{\left(2+\pi\right)^{2}}\right) &= 0\\ \left(q^2+\frac{1}{\left(2+\pi\right)^2}-q^2-p^2\right)m^2+2qpm+p^2+\left(\frac{1}{\left(2+\pi\right)^2}-q^2-p^2\right) &= 0\\ \left(\frac{1}{\left(2+\pi\right)^2}-p^2\right)m^2+2qpm+\frac{1}{\left(2+\pi\right)^2}-q^2 &= 0 \end{align*}$$

Therefore, the possible slopes are given by

$$\begin{align*} m &=\frac{-2qp\pm\sqrt{\left(2qp\right)^2-4\left(\frac{1}{\left(2+\pi\right)^2}-p^2\right)\left(\frac{1}{\left(2+\pi\right)^2}-q^2\right)}}{2\left(\frac{1}{\left(2+\pi\right)^2}-p^2\right)}\\ m &=\frac{-qp\pm\sqrt{q^2p^2-\left(\frac{1}{\left(2+\pi\right)^2}-p^2\right)\left(\frac{1}{\left(2+\pi\right)^2}-q^2\right)}}{\frac{1}{\left(2+\pi\right)^2}-p^2}\\ m &=\frac{-qp\pm\sqrt{q^2p^2-\left(\frac{1}{\left(2+\pi\right)^4}-\frac{p^2+q^2}{\left(2+\pi\right)^2}+p^2q^2\right)}}{\frac{1}{\left(2+\pi\right)^2}-p^2}\\ m &=\frac{-qp\pm\sqrt{q^2p^2-\frac{1}{\left(2+\pi\right)^4}+\frac{p^2+q^2}{\left(2+\pi\right)^2}-p^2q^2}}{\frac{1}{\left(2+\pi\right)^2}-p^2}\\ m &=\frac{-qp\pm\sqrt{\frac{p^2+q^2}{\left(2+\pi\right)^2}-\frac{1}{\left(2+\pi\right)^4}}}{\frac{1}{\left(2+\pi\right)^2}-p^2}\\ m &=\frac{-qp\pm\frac{1}{2+\pi}\sqrt{p^2+q^2-\frac{1}{\left(2+\pi\right)^2}}}{\frac{1}{\left(2+\pi\right)^2}-p^2}\\ m &=\frac{qp\pm\frac{1}{2+\pi}\sqrt{p^2+q^2-\frac{1}{\left(2+\pi\right)^2}}}{p^2-\frac{1}{\left(2+\pi\right)^2}}\\ \end{align*}$$

Because of this, we can use the following equation to give an extra side to $z’$

$$y=\frac{qp+\frac{1}{2+\pi}\sqrt{p^2+q^2-\frac{1}{\left(2+\pi\right)^2}}}{p^2-\frac{1}{\left(2+\pi\right)^{2}}}x$$

This is shown here:

By the same token, we can draw a line from $Z_3$ to a point on $M$ that it is tangent to. We can take our previous equation, and replace $p$ with $p-\frac{\sqrt{5}}{5}$, while replacing $q$ with $q-\frac{\sqrt{5}}{15}$. This equation would be:

$$y-\frac{\sqrt{5}}{15}=\frac{\left(q-\frac{\sqrt{5}}{15}\right)\left(p-\frac{\sqrt{5}}{5}\right)-\frac{1}{2+\pi}\sqrt{\left(p-\frac{\sqrt{5}}{5}\right)^2+\left(q-\frac{\sqrt{5}}{15}\right)^2-\frac{1}{\left(2+\pi\right)^2}}}{\left(p-\frac{\sqrt{5}}{5}\right)^2-\frac{1}{\left(2+\pi\right)^2}}\left(x-\frac{\sqrt{5}}{5}\right)$$

This is shown here:

What is the lowest possible slope of the line tangent to $M$ passing through $Z_1$? We may place the center of $M$ at the point where $f_1$ intersects the circle around $Z_4$. This point is about $\left(0.57391,0.0918346\right)$. Plugging these points in for $p$ and $q$ gives a particular slope for that line. We will define this with $m_1$. Writing out the precise value of $m_1$ would time a very long time, so sufice it to say that $m_1\approx 0.5461546$.

Similarly, I would define $m_2$ as the minimal slope of the line tangent to $M$ passing through $Z_3$. Placing the center of $M$ at the point where $f_1$ intersects the circle around $P$ would mean setting $p\approx 0.115614$, while $q\approx -0.004593$. If $M$ is placed at this point, then $m_2\approx-0.127902662317$.

Now, graph the two lines

$$y=m_1x$$ $$y-\frac{\sqrt{5}}{15}=m_2\left(x-\frac{\sqrt{5}}{5}\right)$$

These two lines intersect at a point that is approximately $(0.306014,0.167131)$. Regardless of where $M$ is located, this point must be included in the convex hull of $M$ and $z’$. Additionally, I’ve been rounding down, meaning that this point really is included, not just one close to it. This extended version of $z’$ is shown below:

Building a taxonomy of type 2 curves

#

I’ve been thinking a lot about $Z$ curves and their variants. I would like to build a framework for thinking about simple type 2 curves. The basic $Z$ curve consists of four points:

$$Z_1=\left(0,0\right)$$ $$Z_2=\left(\frac{1}{3},0\right)$$ $$Z_3=\left(\frac{1}{3},\frac{1}{3}\right)$$ $$Z_4=\left(\frac{2}{3},\frac{1}{3}\right)$$

How might these points be different for other curves? Let’s consider only the unit length 3-chains with parallel end segments. If the basic $Z$ curve is given a variable height $\alpha$, its points may be written as:

$$Z_1=\left(0,0\right)$$ $$Z_2=\left(\frac{1-\alpha}{2},0\right)$$ $$Z_3=\left(\frac{1-\alpha}{2},\alpha\right)$$ $$Z_4=\left(1-\alpha,\alpha\right)$$

It is also possible that the first segment may be lengthened by a distance of $\beta$. Due to the ability to rotate the shape so that the first segment is longer, we may assume $\beta > 0$. Then the points of this curve are:

$$Z_1=\left(0,0\right)$$ $$Z_2=\left(\frac{1-\alpha}{2}+\beta,0\right)$$ $$Z_3=\left(\frac{1-\alpha}{2}+\beta,\alpha\right)$$ $$Z_4=\left(1-\alpha,\alpha\right)$$

Finally, the second segment of this curve does not have to be parallel to the other segments. Let’s shift the third point of this curve to the left or right by a distance of $\gamma$, while letting the fourth point exist where it has to be to maintain a length of 1. Therefore, the length of the second segment is:

$$\sqrt{\alpha^2+\gamma^2}$$

Finally, the locations of the points are:

$$Z_1=\left(0,0\right)$$ $$Z_2=\left(\frac{1-\sqrt{\alpha^2+\gamma^2}}{2}+\beta,0\right)$$ $$Z_3=\left(\frac{1-\sqrt{\alpha^2+\gamma^2}}{2}+\beta+\gamma,\alpha\right)$$ $$Z_4=\left(1-\sqrt{\alpha^2+\gamma^2}+\gamma,\alpha\right)$$

Then, define $Z(\alpha,\beta,\gamma)$ to be the curve defined with these points.

I’m going to go on a brief tangent here, and it’ll be clear soon why this is important. Consider the possible elements of $\left[L,M\right]$ with $M$ and $L$ intersecting. The curve $M$ has a minimum radus of $\frac{1}{\pi+2}$, but it has a maximum radius of $\frac{\sqrt{2}}{\pi+2}$. Let’s locate $L$ on $\left\langle[0,0],[0,1]\right\rangle$. Without loss of generality, let’s assume $M$ is centered above the $x$-axis, and so must have a center with a $y$-coordinate less than $\frac{\sqrt{2}}{\pi+2}$. Additionally, there must not be a point more than 1 unit away from either endpoint of $L$. The possible locations for the center point of $M$ is shown here in black:

The convex hull of these two shapes is shown here in red:

This is a nifty little framework here. Let’s consider elements of $Z(\alpha,0,0)$. These are curves defined with:

$$Z_1=\left(0,0\right)$$ $$Z_2=\left(\frac{1-\alpha}{2},0\right)$$ $$Z_3=\left(\frac{1-\alpha}{2},\alpha\right)$$ $$Z_4=\left(1-\alpha,\alpha\right)$$

An example of these points is shown:

Noteably, this particular element of $\left[L,M\right]$ accomodates $Z(\alpha,0,0)$ as shown! What values of $\alpha$ can be accommodated by any element of $\left[L,M\right]$?

We’ll let $\left(a,b\right)$ be the center of our instance of $M$. For a given value of $\alpha$, which values of $a$ and $b$ accomodate $Z_4$? Assuming $Z$ is located as in the example, $Z_4$ lies on the line $y=-x+1$, while $Z_3$ lies on the line $y=-2x+1$. The line defining the top left boundary of the red region is given by:

$$y=\frac{ba+u_{1}}{v}x$$

Where $v=\left(a^{2}-r^{2}\right)$ and $u_{1}=r\sqrt{b^{2}+v}$. Let’s assume that the slope of this line is at a minimum, meaning that $a=1-r$ and $b=0$. In that case, the line defining the top left boundary is:

$$y=\frac{r\sqrt{v}}{v}x = \frac{r}{\sqrt{v}}x= \frac{r}{\sqrt{1-2r}}x$$

This means that $Z_3$ is only just accomodated when these lines intersect at

$$\begin{align*} -2x+1 &= \frac{r}{\sqrt{1-2r}}x\\ 1 &= \left(2+\frac{r}{\sqrt{1-2r}}\right)x\\ 1 &= \left(\frac{2{\sqrt{1-2r}}}{\sqrt{1-2r}}+\frac{r}{\sqrt{1-2r}}\right)x\\ 1 &= \left(\frac{r+2{\sqrt{1-2r}}}{\sqrt{1-2r}}\right)x\\ x &= \frac{\sqrt{1-2r}}{r+2{\sqrt{1-2r}}}\\ \end{align*}$$

At this point, the $y$-coordinate is given by:

$$\begin{align*} \frac{-2\sqrt{1-2r}}{r+2{\sqrt{1-2r}}}+1 &= \frac{-2\sqrt{1-2r}}{r+2{\sqrt{1-2r}}}+\frac{r+2{\sqrt{1-2r}}}{r+2{\sqrt{1-2r}}}\\ &= \frac{-2\sqrt{1-2r}+r+2{\sqrt{1-2r}}}{r+2{\sqrt{1-2r}}}\\ &= \frac{r}{r+2{\sqrt{1-2r}}}\\ \end{align*}$$

Therefore, $Z_3$ is accomodated whenever $\alpha < \frac{r}{r+2{\sqrt{1-2r}}}$. When is $Z_4$ accomodated? The top left boundary of the red zone is given by

$y=\frac{b-ab+u_2}{-p}\left(x-1\right)$

Where $p=v-2a+1$ and $u_2=r\sqrt{b^2+p}$. Because $Z_4$ is located on $y=-x+1$, $Z_4$ is accomodated whenever:

$-x+1<\frac{b-ab+u_2}{-p}\left(x-1\right)$

Rather than go through all the effort of solving this equation, I’ll simply point out that the lowest possible location for $\left(a,b\right)$ to be while accomodating $Z_4$ is the one where $M$ is tangent to $y=-x+1$. This equation is:

$y=-x+1-\sqrt{2}r$

The reduced area for $(a,b)$ is shown below:

Next, let’s consider rotating $Z$ so that $Z_2$ is located on the $x$-axis and $Z_4$ is located at $(1,0)$. The distance between $Z_2$ and $Z_4$ is given by:

$$\begin{align*} \sqrt{\alpha^2+\left(\frac{1-\alpha}{2}\right)^2} &= \sqrt{\alpha^2+\frac{1-2\alpha+\alpha^2}{4}}\\ &= \sqrt{\frac{1-2\alpha+5\alpha^2}{4}}\\ &= \frac{\sqrt{1-2\alpha+5\alpha^2}}{2}\\ \end{align*}$$

This places $Z_2$ at $\left(1-\frac{\sqrt{1-2\alpha+5\alpha^2}}{2},0\right)$. To locate $Z_3$, we’ll find the slopes from $Z_3$ to $Z_2$ and $Z_4$, and set them perpendicular to each other:

$$\begin{align*} \frac{1-x}{y}&= \frac{y}{x-1+\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2}}\\ (1-x)\left(x-1+\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2}\right)&= y^2\\ \end{align*}$$

We also have that it must be a distance of $\frac{1-\alpha}{2}$ from $Z_4$. This gives the limitation that

$$\begin{align*} y^2 +(x-1)^2 &= \left(\frac{1-\alpha}{2}\right)^2\\ y^2 &= \left(\frac{1-\alpha}{2}\right)^2-(x-1)^2\\ \end{align*}$$

Next, we may combine these and solve for $x$.

$$\begin{align*} (1-x)\left(x-1+\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2}\right)&= \left(\frac{1-\alpha}{2}\right)^2-(x-1)^2\\ -(x-1)\left((x-1)+\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2}\right)&= \left(\frac{1-\alpha}{2}\right)^2-(x-1)^2\\ (1-x)\left(\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2}\right)&= \left(\frac{1-\alpha}{2}\right)^2\\ (1-x)\left(\sqrt{1-2\alpha+5\alpha^{2}}\right)&= \frac{\left(1-\alpha\right)^2}{2}\\ (1-x)&= \frac{\left(1-\alpha\right)^2}{2\sqrt{1-2\alpha+5\alpha^{2}}}\\ 1-x&= \frac{\left(1-\alpha\right)^2}{2\sqrt{1-2\alpha+5\alpha^{2}}}\\ x&= 1-\frac{\left(1-\alpha\right)^2}{2\sqrt{1-2\alpha+5\alpha^{2}}}\\ \end{align*}$$

Plugging this in lets us find $y$ as well:

$$\begin{align*} y^2 &= \left(\frac{1-\alpha}{2}\right)^2-\left(\frac{\left(1-\alpha\right)^2}{2\sqrt{1-2\alpha+5\alpha^{2}}}\right)^2\\ y^2 &= \frac{\left(1-\alpha\right)^2}{4}-\frac{\left(1-\alpha\right)^4}{4\left(1-2\alpha+5\alpha^2\right)}\\ y^2 &= \frac{\left(1-\alpha\right)^2}{4}\left(1-\frac{\left(1-\alpha\right)^2}{\left(1-2\alpha+5\alpha^2\right)}\right)\\ y^2 &= \frac{\left(1-\alpha\right)^2}{4}\left(\frac{1-2\alpha+5\alpha^2}{1-2\alpha+5\alpha^2}-\frac{1-2\alpha + \alpha^2}{1-2\alpha+5\alpha^2}\right)\\ y^2 &= \frac{\left(1-\alpha\right)^2}{4}\left(\frac{4\alpha^2}{1-2\alpha+5\alpha^2}\right)\\ y^2 &= \frac{\alpha^2\left(1-\alpha\right)^2}{1-2\alpha+5\alpha^2}\\ y &= \frac{\alpha(1-\alpha)}{\sqrt{1-2\alpha+5\alpha^2}}\\ \end{align*}$$

Therefore, $Z_3=\left(1-\frac{\left(1-\alpha\right)^2}{2\sqrt{1-2\alpha+5\alpha^{2}}},\frac{\alpha(1-\alpha)}{\sqrt{1-2\alpha+5\alpha^2}}\right)$

Finally, finding $Z_1$ is fairly straightforward:

$$\begin{align*} Z_1 &= \left(1-\frac{\left(1-\alpha\right)^2}{2\sqrt{1-2\alpha+5\alpha^{2}}}-\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2},\frac{\alpha(1-\alpha)}{\sqrt{1-2\alpha+5\alpha^2}}\right)\\ Z_1 &= \left(1-\frac{1-2\alpha+\alpha^2}{2\sqrt{1-2\alpha+5\alpha^{2}}}-\frac{1-2\alpha+5\alpha^{2}}{2\sqrt{1-2\alpha+5\alpha^{2}}},\frac{\alpha(1-\alpha)}{\sqrt{1-2\alpha+5\alpha^2}}\right)\\ Z_1 &= \left(1+\frac{-2+4\alpha-6\alpha^2}{2\sqrt{1-2\alpha+5\alpha^{2}}},\frac{\alpha(1-\alpha)}{\sqrt{1-2\alpha+5\alpha^2}}\right)\\ Z_1 &= \left(1-\frac{1-2\alpha-3\alpha^2}{\sqrt{1-2\alpha+5\alpha^{2}}},\frac{\alpha(1-\alpha)}{\sqrt{1-2\alpha+5\alpha^2}}\right)\\ \end{align*}$$

Finally, Here’s that rotated version displayed:

The line containing $Z_1$ is given by:

$y=\frac{\alpha\left(1-\alpha\right)}{\sqrt{1-2\alpha+5\alpha^{2}}-1+2\alpha-3\alpha^{2}}x$

Once again, the parallel line that is located a distance of $r$ under the previous one is:

$y+\frac{r}{\sqrt{m^2+1}}=m\left(x-\frac{mr}{\sqrt{m^2+1}}\right)$

where $m$ is the slope from the previous equation. Because the center of $M$ must be located underneath this, the total possible area for $(a,b)$ is reduced even further.

Next, let’s rotate $Z$ to have both end points on the $x$ axis. The distance between $Z_1$ and $Z_4$ is given by:

$\sqrt{\alpha^2+(1-\alpha)^2}=\sqrt{\alpha^2+1-2\alpha+\alpha^2}=\sqrt{2\alpha^2-2\alpha+1}$

As before, the distance from $Z_1$ to $Z_2$ is $\frac{1-\alpha}{2}$, while the distance from $Z_4$ to $Z_2$ is $\frac{\sqrt{1-2\alpha+5\alpha^2}}{2}$. Therefore, if $Z_1$ is located at the origin, then the $x$ and $y$-coordinates of $Z_2$ are restricted by:

$y^2+x^2=\left(\frac{1-\alpha}{2}\right)^2$ $y^2+\left(x-\sqrt{2\alpha^2-2\alpha+1}\right)^2=\left(\frac{\sqrt{1-2\alpha+5\alpha^2}}{2}\right)^2$ $y<0$

Setting $y^2$ equal to each other and solving for $x$ gives:

$$\begin{align*} \left(\frac{1-\alpha}{2}\right)^{2}-x^{2} &= \left(\frac{\sqrt{1-2\alpha+5\alpha^{2}}}{2}\right)^{2}-\left(x-\sqrt{2\alpha^{2}-2\alpha+1}\right)^{2}\\ \frac{\left(1-\alpha\right)^{2}}{4}-x^{2} &= \frac{1-2\alpha+5\alpha^{2}}{4}-\left(x^{2}-2x\sqrt{2\alpha^{2}-2\alpha+1}+2\alpha^{2}-2\alpha+1\right)\\ \frac{\left(1-\alpha\right)^{2}}{4} &= \frac{1-2\alpha+5\alpha^{2}}{4}+2x\sqrt{2\alpha^{2}-2\alpha+1}-2\alpha^{2}+2\alpha-1\\ 0 &= \frac{4\alpha^{2}}{4}+2x\sqrt{2\alpha^{2}-2\alpha+1}-2\alpha^{2}+2\alpha-1\\ 0 &= 2x\sqrt{2\alpha^{2}-2\alpha+1}-\alpha^{2}+2\alpha-1\\ \left(\alpha-1\right)^{2} &= 2x\sqrt{2\alpha^{2}-2\alpha+1}\\ x &= \frac{\left(\alpha-1\right)^{2}}{2\sqrt{2\alpha^{2}-2\alpha+1}}\\ \end{align*}$$

Then, solving for $y$ gives:

$$\begin{align*} y^{2} &= \frac{\left(\alpha-1\right)^{2}}{4}-\frac{\left(\alpha-1\right)^{4}}{4\left(2\alpha^{2}-2\alpha+1\right)}\\ y^{2} &= \left(\alpha-1\right)^{2}\left(\frac{\left(2\alpha^{2}-2\alpha+1\right)}{4\left(2\alpha^{2}-2\alpha+1\right)}-\frac{\left(\alpha-1\right)^{2}}{4\left(2\alpha^{2}-2\alpha+1\right)}\right)\\ y^{2} &= \left(\alpha-1\right)^{2}\left(\frac{\left(2\alpha^{2}-2\alpha+1\right)-\left(\alpha-1\right)^{2}}{4\left(2\alpha^{2}-2\alpha+1\right)}\right)\\ y^{2} &= \left(\alpha-1\right)^{2}\left(\frac{\alpha^{2}}{4\left(2\alpha^{2}-2\alpha+1\right)}\right)\\ y &= \frac{\alpha\left(\alpha-1\right)}{2\sqrt{2\alpha^{2}-2\alpha+1}}\\ \end{align*}$$

Finally, the points within $Z$ are given by:

$$\begin{align*} Z_1 &= \left(0,0\right)\\ Z_2 &= \left(\frac{\left(\alpha-1\right)^{2}}{2\sqrt{2\alpha^{2}-2\alpha+1}},\frac{\alpha\left(\alpha-1\right)}{2\sqrt{2\alpha^{2}-2\alpha+1}}\right)\\ Z_3 &= \left(\frac{3\alpha^{2}-2\alpha+1}{2\sqrt{2\alpha^{2}-2\alpha+1}},\frac{\alpha\left(1-\alpha\right)}{2\sqrt{2\alpha^{2}-2\alpha+1}}\right)\\ Z_4 &= \left(\sqrt{2\alpha^{2}-2\alpha+1},0\right)\\ \end{align*}$$

Up next, I want to shift this instance of $Z$ upwards as much as possible. As mentioned, the slope of the upper left edge of the red zone is:

$$\frac{ba+u_{1}}{v}$$

Let’s say we want to shift $Z$ up by $y_1$ and to the right by $x_1$. If it’s up as high as possible, we have that

$$y_1=\frac{ba+u_{1}}{v} x_1$$

The location of $Z_4$ is now $\left(\sqrt{2\alpha^{2}-2\alpha+1} + x_1, y_1\right)$

And this must lie on or below the red zone’s upper right boundary of

$$y = \frac{b-ab+u_{2}}{-p}\left(x-1\right)$$

plugging in the location of $Z_4$ gives a requirement of

$$y_1 = \frac{b-ab+u_{2}}{-p}\left(\sqrt{2\alpha^{2}-2\alpha+1} + x_1-1\right)$$

And plugging in the requirement from a few lines ago gives:

$$\begin{align*} \frac{ab+u_{1}}{v} x_1 &= \frac{b-ab+u_{2}}{-p}\left(\sqrt{2\alpha^{2}-2\alpha+1} + x_1-1\right)\\ \left(\frac{ab+u_{1}}{v}+\frac{b-ab+u_{2}}{p}\right) x_1 &= \frac{b-ab+u_{2}}{-p}\left(\sqrt{2\alpha^{2}-2\alpha+1} + -1\right)\\ \left(\frac{abp+u_1p}{vp}+\frac{bv-abv+u_2v}{vp}\right) x_1 &= \frac{b-ab+u_2}{p}\left(1-\sqrt{2\alpha^2-2\alpha+1}\right)\\ \left(\frac{abp+u_1p+bv-abv+u_2v}{v}\right) x_1 &= \left(b-ab+u_2\right)\left(1-\sqrt{2\alpha^2-2\alpha+1}\right)\\ x_1 &= \frac{v\left(q+u_2\right)\left(1-\sqrt{2\alpha^2-2\alpha+1}\right)}{abp+u_1p+qv+u_2v}\\ \end{align*}$$

And then we also get that:

$$y_1 &= \frac{(ba+u_1)\left(q+u_2\right)\left(1-\sqrt{2\alpha^2-2\alpha+1}\right)}{abp+u_1p+qv+u_2v}$$

This is shown below:

When does $M$ contain $Z_2$? This is the only point that could possibly be excluded. Well, the bottom half of $M$ is defined by:

$$y=-\sqrt{r^2-(x-a)^2}+b$$

The location of $Z_2$ is given by:

$$\left(\frac{\left(\alpha-1\right)^{2}}{2\sqrt{2\alpha^{2}-2\alpha+1}}+x_1,\frac{\alpha\left(\alpha-1\right)}{2\sqrt{2\alpha^{2}-2\alpha+1}}+y_1\right)$$

Therefore, $Z_2$ is inside $M$ whenever:

$$\frac{\alpha\left(\alpha-1\right)}{2\sqrt{2\alpha^{2}-2\alpha+1}}+y_1>-\sqrt{r^2-\left(\frac{\left(\alpha-1\right)^{2}}{2\sqrt{2\alpha^{2}-2\alpha+1}}+x_1-a\right)^2}+b$$

Trying to simplify this mess would be a disaster. Instead, I’ll just