Continuing Analysis of Mosers worm

Introduction

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This is the second half of a project introduced earlier on this website! To see that, look here:

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 question to explore, 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:

Placing M and G on the plane

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I’m curious if it’s possible to accommodate \(Z(h, 0, 0)\) with \(G\) and \(M\). To this end, let \(c=\left(c_x,c_y\right)\) be the center of a circle \(m\) with radius \(r=\frac{1}{\pi+2}\), and let \(g\) be located along the non-negative \(x\)-axis, starting at the origin. I’ll also define \(d=\sqrt{c_x^2+c_y^2}\) to be the distance from \(c\) to the origin, and define \(\theta_c=\arctan\frac{c_y}{c_x}\) to be the angle to \(c\), so that \(c\) can be expressed in polar coordinates as \(\left(d,\theta_c\right)\). We may also define \(\theta_d = \arcsin\left(\frac{r}{d}\right)\) so that the two lines tangent to \(m\) passing through the origin are at angles of \(\theta_c+\theta_d\) and \(\theta_c-\theta_d\), and have lengths of \(\sqrt{d^2-r^2}\). Finally, define \(z_h=\left\langle Z_1, Z_2, Z_3, Z_4\right\rangle\) to be an instance of \(Z(h,0,0)\).

Along the same lines, we can allow \(d’=\sqrt{\left(1-c_x\right)^2+c_y^2}\) to be the distance from \(c\) to \(\left(1,0\right)\), and let\(\theta_c’=\arctan\frac{c_y}{c_x-1}\) be the angle formed by \(c\) and \(\left(1,0\right)\). Similarly, let’s define \(\theta_d’ = \arcsin\left(\frac{r}{d’}\right)\), so that the lines passing through \(\left(1,0\right)\) tangent to \(m\) are at angles of \(\theta_c’+\theta_d’\) and \(\theta_c’-\theta_d’\), and have lengths of \(\sqrt{d’^2-r^2}\).

In order to keep things straight, let’s also name the straight lines \(t_1\), \(t_2\), \(t_3\), and \(t_4\), as shown:

Suppose we wanted to place \(z_h\) with the first point at the origin. Let’s define the angle \(\alpha\) to be the angle of the second point in \(z_h\). We may define \(\zeta_3 = \tan\left(\frac{2h}{1-h}\right)\) and \(\zeta_4 = \arctan\left(\frac{h}{1-h}\right)\) to be the angles of the third and fourth points in this collection, and define \(\r_2=\frac{1-h}{2}\), \(\r_3=\frac{\sqrt{1-2h+5h^2}}{2}\), and \(\r_4=\sqrt{1-2h+2h^2}\) to be the distances of the second, third, and fourth points from the origin. The locations of the points in \(z_h\) are defined in polar coordinates with:

$$Z_1=\left(0,0\right)$$ $$Z_2=\left(r_2,\alpha\right)$$ $$Z_3=\left(r_3, \alpha+\zeta_3\right)$$ $$Z_4=\left(r_4, \alpha+\zeta_4\right)$$

For convenience, I’ll reiterate all the definitions here:

Finally, everything has been defined, and things can finally get interesting. For this section, let’s assume that \(m\) and \(g\) are fully intersecting. Without loss of generality, we may assume that \(c\) is above the \(x\)-axis. Since \(m\) has a radius of \(r\), we may conclude that \(c_y\in [0,r]\). Additionally, we know that no point may be more than one unit away from another, so \(c\) must be within a distance of \(1-r\) from both the origin and \((1,0)\). The possible locations of \(c\) are shown here in black:

Defining and restricting two points

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I am going to consider three specific positions of \(z_h\). For now, I’ll restrict our considerations to the first two. They are:

1. \(Z_1\) is located at the origin, \(Z_2\) lies on \(t_2\)
2. \(Z_1\) is located at the \((1,0)\), \(Z_3\) lies on \(t_4\)

For both of these positions, it is likely that \(Z_2\) and \(Z_3\) lie inside \([m,g]\), but it is not guaranteed for \(Z_4\). Therefore, define these points as follows:

1. \(P_1=Z_4\) when \(Z_1\) is located at the origin, \(Z_2\) lies on \(t_2\)
2. \(P_2=Z_4\) when \(Z_1\) is located at the \((1,0)\), \(Z_3\) lies on \(t_4\)

Requirements for \(P_1\) and \(P_2\)

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We want to find a set of requirements that guarantee that these points exist. I’ll focus on \(P_1\) for a while. In order to guarantee that \(Z_2\) and \(Z_3\) lie inside \([m,g]\), the existence of these two points will come with the following requirements[2]I’ll note that these requirements are sufficient, but not neccesary to guarantee that \(Z_2\),\(Z_3 \in [m,g]\):

• \(r_3 \leq \sqrt{d^2-r^2}\)
• \(\zeta_3 \leq 2\theta_d\)

Immediately, we have that

$$r_3 \leq \sqrt{d^2-r^2} \Rrightarrow d \geq \sqrt{r_3^2+r^2}$$

Satisfying the second requirement is a bit tougher. Substituting in our definitions for these terms, we have

$$\arccos\frac{1-h}{2r_3} \leq 2\arcsin\frac{r}{d}$$

A little bit of trig substitution lets us solve for \(d\):

$$\begin{align*} \frac{r}{d} &\geq \sin{\left(\frac{1}{2}\arccos\frac{1-h}{2r_3}\right)} = \sqrt{\frac{1-\cos{\left(\arccos\frac{1-h}{2r_3}\right)}}{2}}\\ &= \sqrt{\frac{1-\frac{1-h}{2r_3}}{2}} = \sqrt{\frac{2r_3+h-1}{4r_3}}\\ \frac{d}{r} &\leq \sqrt{\frac{4r_3}{2r_3+h-1}}\\ d &\leq r\sqrt{\frac{4r_3}{2r_3+h-1}}\\ \end{align*}$$

Taken together, we’ll say that \(P_1\) only exists when \(d\in\left[\sqrt{r_3^2+r^2},r\sqrt{\frac{4r_3}{2r_3+h-1}}\right]\). Identical reasoning tells us that \(P_2\) only exists when \(d’\in\left[\sqrt{r_3^2+r^2},r\sqrt{\frac{4r_3}{2r_3+h-1}}\right]\).

The extrema of the ordered pair \(h,d\) under the limitation that \(d = \sqrt{r_3^2+r^2}\) are:

$$\left(0, \frac{\sqrt{1+4r^2}}{2}\right)\approx\left(0,0.536\right)$$ $$\left(\frac{1}{5}, \sqrt{0.2+r^2}\right)\approx\left(0.2,0.488\right)$$ $$\left(\frac{1}{3}, \sqrt{2/9+r^2}\right)\approx\left(0.333,0.510\right)$$

Notably, all these values of \(d\) are fairly close together, meaning that \(h\) does not dramatically affect the lower bound of \(d\). The upper bound of \(d\) is much more reliant on \(h\), but it is greater than \(1-r\) whenever \(h\) is less than about 0.21. An example of this range is shown, where \(h=0.2\)

Locating \(P_1\) and \(P_2\)

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While locating \(P_1\), we wish for \(Z_2\) to lie on \(t_2\). To achieve this, we’ll set \(\alpha = \theta_c - \theta_d\). Naturally, the location of \(Z_4\) (and by extension \(P_1\)) is:

$$\left(r_4\cos\left(\theta_c - \theta_d + \zeta_4\right),r_4\sin\left(\theta_c - \theta_d + \zeta_4\right)\right)$$

This point is sometimes located within \([m,g]\) for some locations of \(c\), but not all. An example of each is shown:

Using some trig identities, The location of \(P_1\) can be rewritten in terms of \(\theta_c - \theta_d\) and \(\zeta_4\):

$$\left(r_4 \left(\cos{\left(\theta_c - \theta_d\right)}\cos{\zeta_4} - \sin{\left(\theta_c - \theta_d\right)}\sin{\zeta_4}\right),r_4 \left(\sin{\left(\theta_c - \theta_d\right)}\cos{\zeta_4}+\cos{\left(\theta_c - \theta_d\right)}\sin{\zeta_4}\right)\right)$$

A little more work will rewrite this in terms of \(c_x\) and \(c_y\). Again, we can use trig substitutions to get

$$\begin{align*} \sin{\left(\theta_c - \theta_d\right)} &= \sin{\theta_c}\cos{\theta_d}-\sin{\theta_d}\cos{\theta_c}\\ &= \sin{\left(\arctan\frac{c_y}{c_x}\right)}\cos{\left(\arcsin\frac{r}{d}\right)}-\sin{\left(\arcsin\frac{r}{d}\right)}\cos{\left(\arctan\frac{c_y}{c_x}\right)}\\ &= \frac{c_y}{d}\cdot\frac{\sqrt{d^2-r^2}}{d}-\frac{r}{d}\cdot\frac{c_x}{d}\\ &= \frac{kc_y-rc_x}{d^2} \end{align*}$$

Where \(k=\sqrt{d^2-r^2}\). Similarly, we also have:

$$\begin{align*} \cos{\left(\theta_c - \theta_d\right)} &= \cos{\theta_c}\cos{\theta_d}+\sin{\theta_d}\sin{\theta_c}\\ &= \frac{c_x}{d}\frac{k}{d}+\frac{r}{d}\frac{c_y}{d}\\ &= \frac{kc_x+rc_y}{d^2} \end{align*}$$

Therefore, the position of \(P_1\) can also be written as

$$\left(r_4 \left(\frac{kc_x+rc_y}{d^2}\cdot\cos{\zeta_4} - \frac{kc_y-rc_x}{d^2}\cdot\sin{\zeta_4}\right),r_4 \left(\frac{kc_y-rc_x}{d^2}\cdot\cos{\zeta_4}+\frac{kc_x+rc_y}{d^2}\cdot\sin{\zeta_4}\right)\right)$$

Finally, we may also note that \(\cos{\zeta_4} = \frac{1-h}{r_4}\) and \(\sin{\zeta_4} = \frac{h}{r_4}\). This allows us to rewrite \(P_1\) once again as

$$\left((1-h)\frac{kc_x+rc_y}{d^2} - h\frac{kc_y-rc_x}{d^2},(1-h)\frac{kc_y-rc_x}{d^2}+h\frac{kc_x+rc_y}{d^2}\right)$$

The location of \(P_2\) can be found with the same method. The ordinary way to locate it would be:

$$\left(1+r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right),r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)\right)$$

And just like before, we’ll use the angle sum identities to rewrite this:

$$r_4\left(\frac{1}{r_4}+\cos\left(\theta_c’+\theta_d’\right)\cos\left(\zeta_4-\zeta_3\right)-\sin\left(\theta_c’+\theta_d’\right)\sin\left(\zeta_4-\zeta_3\right), \sin\left(\theta_c’+\theta_d’\right)\cos\left(\zeta_4-\zeta_3\right)+\cos\left(\theta_c’+\theta_d’\right)\sin\left(\zeta_4-\zeta_3\right)\right)$$

Next, we’ll rewrite \(\cos\left(\theta_c’+\theta_d’\right)\) and \(\sin\left(\theta_c’+\theta_d’\right)\):

$$\begin{align*} \cos\left(\theta_c’+\theta_d’\right) &= \cos\left(\theta_c’\right)\cos\left(\theta_d’\right)-\sin\left(\theta_c’\right)\sin\left(\theta_d’\right)\\ &= -\cos\left(\arctan\frac{c_y}{c_x-1}\right)\cos\left(\arcsin\frac{r}{d’}\right)+\sin\left(\arctan\frac{c_y}{c_x-1}\right)\sin\left(\arcsin\frac{r}{d’}\right)\\ &= -\frac{1}{\sqrt{1+\frac{c_y^2}{(c_x-1)^2}}}\sqrt{1-\frac{r^2}{d’^2}}+\frac{\frac{c_y}{c_x-1}}{\sqrt{1+\frac{c_y^2}{(c_x-1)^2}}}\frac{r}{d’}\\ &= -\sqrt{\frac{(c_x-1)^2}{(c_x-1)^2+c_y^2}}\cdot\sqrt{1-\frac{r^2}{d’^2}}+\frac{c_y}{c_x-1}\cdot\sqrt{\frac{(c_x-1)^2}{(c_x-1)^2+c_y^2}}\cdot\frac{r}{d’}\\ &= -\frac{c_x-1}{d’}\cdot\sqrt{\frac{d’^2-r^2}{d’^2}}+\frac{c_y}{c_x-1}\cdot\frac{c_x-1}{d’}\cdot\frac{r}{d’}\\ &= \frac{c_y}{c_x-1}\cdot\frac{r(c_x-1)}{d’^2}-\frac{k’(c_x-1)}{d’^2}\\ &= \frac{rc_y-k’(c_x-1)}{d’^2}\\ \end{align*}$$

Where \(k’=\sqrt{d’^2-r^2}\). Similarly, we also have:

$$\begin{align*} \sin\left(\theta_c’+\theta_d’\right) &= \sin\left(\theta_c’\right)\cos\left(\theta_d’\right)+\cos\left(\theta_c’\right)\sin\left(\theta_d’\right)\\ &= -\sin\left(\arctan\frac{c_y}{c_x-1}\right)\cos\left(\arcsin\frac{r}{d’}\right)-\cos\left(\arctan\frac{c_y}{c_x-1}\right)\sin\left(\arcsin\frac{r}{d’}\right)\\ &= -\frac{\frac{c_y}{c_x-1}}{\sqrt{1+\frac{c_y^2}{(c_x-1)^2}}}\sqrt{1-\frac{r^2}{d’^2}}-\frac{1}{\sqrt{1+\frac{c_y^2}{(c_x-1)^2}}}\cdot\frac{r}{d’}\\ &= -\frac{c_y}{c_x-1}\cdot\sqrt{\frac{(c_x-1)^2}{(c_x-1)^2+c_y^2}}\cdot\sqrt{1-\frac{r^2}{d’^2}}-\sqrt{\frac{(c_x-1)^2}{(c_x-1)^2+c_y^2}}\cdot\frac{r}{d’}\\ &= -\frac{c_y}{c_x-1}\cdot\frac{c_x-1}{d’}\cdot\sqrt{1-\frac{r^2}{d’^2}}-\frac{c_x-1}{d’}\cdot\frac{r}{d’}\\ &= -\frac{k’c_y}{d’^2}-\frac{r(c_x-1)}{d’^2}\\ &= \frac{-r(c_x-1)-k’c_y}{d’^2}\\ \end{align*}$$

Just like before, \(\cos{\zeta_4} = \frac{1-h}{r_4}\) and \(\sin{\zeta_4} = \frac{h}{r_4}\). Additionally, \(\cos{\zeta_3} = \frac{1-h}{2r_3}\) and \(\sin{\zeta_3} = \frac{h}{r_3}\), which allows for:

$$\begin{align*} \cos\left(\zeta_4-\zeta_3\right) &= \cos\left(\zeta_4\right)\cos\left(\zeta_3\right)+\sin\left(\zeta_4\right)\sin\left(\zeta_3\right)\\ &= \frac{1-h}{r_4}\cdot\frac{1-h}{2r_3}+\frac{h}{r_4}\cdot\frac{h}{r_3}\\ &= \frac{(1-h)^2}{2r_3r_4}+\frac{2h^2}{2r_3r_4} = \frac{3h^2-2h+1}{2r_3r_4}\\ \end{align*}$$

and

$$\begin{align*} \sin\left(\zeta_4-\zeta_3\right) &= \sin\left(\zeta_4\right)\cos\left(\zeta_3\right)-\cos\left(\zeta_4\right)\sin\left(\zeta_3\right)\\ &= \frac{h}{r_4}\cdot\frac{1-h}{2r_3}-\frac{1-h}{r_4}\cdot\frac{h}{r_3}\\ &= \frac{(1-h)h}{2r_3r_4}-\frac{2(1-h)h}{2r_3r_4}\\ &= -\frac{(1-h)h}{2r_3r_4}=\frac{h^2-h}{2r_3r_4}\\ \end{align*}$$

Just like before, we can give the updated version of \(P_2\):

$$\left(1+\frac{\left(rc_y-k’(c_x-1)\right)\left(3h^2-2h+1\right)+\left(r(c_x-1)+k’c_y\right)(h^2-h)}{2r_3d’^2},\ \frac{\left(rc_y-k’(c_x-1)\right)(h^2-h)-\left(r(c_x-1)+k’c_y\right)\left(3h^2-2h+1\right)}{2r_3d’^2}\right)$$

Although, admittedly, this “updated” version isn’t very short or clean compared to the original. Instead, the location of \(P_2\) in polar coordinates from \((1,0)\) is:

$$\left(r_4, \theta_c’+\theta_d’+\zeta_4-\zeta_3\right)$$

Using, \(\beta=\theta_c’+\theta_d’+\zeta_4-\zeta_3\), then the distance of this point to the origin is:

$$\begin{align*} \ell\left\langle(0,0),Z_4\right\rangle &= \sqrt{\left(r_4\cos\left(\beta\right)+1\right)^2+\left(r_4\sin\left(\beta\right)\right)^2}\\ &= \sqrt{r_4^2\cos^2\left(\beta\right)+2r_4\cos\left(\beta\right)+1+r_4^2\sin^2\left(\beta\right)}\\ &= \sqrt{r_4^2+2r_4\cos\left(\beta\right)+1}\\ \end{align*}$$

Then the angle to \(Z_4\) from the origin is found with the cosine law:

$$\arccos\left(\frac{1+r_4^2+2r_4\cos\left(\beta\right)+1-r_4^2}{2\sqrt{r_4^2+2r_4\cos\left(\beta\right)+1}}\right) = \arccos\left(\frac{1+r_4\cos\left(\beta\right)}{\sqrt{r_4^2+2r_4\cos\left(\beta\right)+1}}\right)$$

So the location of of \(P_2\) is found in polar coorinates with:

$$P_2=\left(\sqrt{r_4^2+2r_4\cos\left(\beta\right)+1},\arccos\left(\frac{1+r_4\cos\left(\beta\right)}{\sqrt{r_4^2+2r_4\cos\left(\beta\right)+1}}\right)\right)$$

Or, alternately

$$P_2=\left(\sqrt{r_4^2+2r_4\cos\left(\beta\right)+1},\arctan\left(\frac{r_4\sin\left(\beta\right)}{r_4\cos\left(\beta\right)+1}\right)\right)$$

Restricting the location of \(c\)

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All things considered, the two points described are:

\(P_1\):

$$ \begin{array}{| c | c |} \hline \text{Location in cartesian} & \left((1-h)\frac{kc_x+rc_y}{d^2} - h\frac{kc_y-rc_x}{d^2},(1-h)\frac{kc_y-rc_x}{d^2}+h\frac{kc_x+rc_y}{d^2}\right) \\ \hline \text{Location in polar} & \left(r_4,\theta_c-\theta_d+\zeta_4\right) \\ \hline \text{Requirements} & d\in\left[\sqrt{r_3^2+r^2},r\sqrt{\frac{4r_3}{2r_3+h-1}}\right] \\ \hline \end{array} $$

\(P_2\):

$$ \begin{array}{| c | c |} \hline \text{Location in cartesian} & \left(1+r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right),r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)\right) \\ \hline \text{Location in polar from origin} & \left(\sqrt{r_4^2+2r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)+1},\arctan\left(\frac{r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)}{r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)+1}\right)\right) \\ \hline \text{Location in polar from }(1,0) & \left(r_4, \theta_c’+\theta_d’+\zeta_4-\zeta_3\right) \\ \hline \text{Requirements} & d’\in\left[\sqrt{r_3^2+r^2},r\sqrt{\frac{4r_3}{2r_3+h-1}}\right] \\ \hline \end{array} $$

If we want to place \(z_h\) inside \([m,g]\), we need to find the possible locations of \(c\) that does not place \(P_1\) or \(P_2\) inside \([m,g]\). I’ll note that it is possible for \(P_1\) to be located underneath \(t_4\). However, for any position of \(c\) that permits \(P_1\) to exist, it is also possible to place \(Z_1\) on the origin at a particular angle that puts \(Z_2\) and \(Z_3\) inside \([m,g]\) with \(Z_4\) above \(t_3\). Therefore, if \(P_1\) exists and is underneath \(t_3\), I’ll consider that location of \(c\) to be guaranteed to allow \(z_h\) inside \([m,g]\).

In addition, I’ll note that the range of possible values of \(d\) disappear almost completely for values of \(h\) above around \(0.3\). It’s worth mentioning that in these cases, it is largely possible to shift the position of \(z_h\) slightly in order to keep \(Z_1\), \(Z_2\), and \(Z_3\) inside \([m,g\), meaning that these limitations can be left aside. In the interest of brevity, and in order to create a version of this project that is complete rather than perfect, I’ll assume that \(h\leq 0.2\), meaning that the requirements for \(P_1\) are at their most generous.

There are two possible ways to show when \(P_1\) lies under \(t_3\). The first is by considering the location of \(P_1\) in cartesian coordinates, and the second is by considering its location in polar coordinates.

Requirements with cartesian coordinates:

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To find the values of \(c_x\) and \(c_y\) that place \(P_1\) below \(t_3\), we place \(c_x\) and \(c_y\) into the equation \(y\leq \tan{\left(\theta_c’-\theta_d’\right)}(x-1)\):

$$r_4 \left(\frac{kc_y-rc_x}{d^2}\cdot\cos{\zeta_4}+\frac{kc_x+rc_y}{d^2}\cdot\sin{\zeta_4}\right)\leq \tan{\left(\theta_c’-\theta_d’\right)}(r_4 \left(\frac{kc_x+rc_y}{d^2}\cdot\cos{\zeta_4} - \frac{kc_y-rc_x}{d^2}\cdot\sin{\zeta_4}\right)-1)$$

We can also rewrite \(\tan{\left(\theta_c’-\theta_d’\right)}\) using trig substitutions:

$$\begin{align*} \tan{\left(\theta_c’-\theta_d’\right)} &= \frac{\tan{\theta_c’}-\tan{\theta_d’}}{1+\tan{\theta_c’}\tan{\theta_d’}}\\ &= \frac{\tan{\theta_c’}-\tan{\theta_d’}}{1+\tan{\theta_c’}\tan{\theta_d’}}\\ &= \frac{\tan{\left(\arctan\frac{c_y}{c_x-1}\right)}-\tan{\left(\arcsin\frac{r}{d’}\right)}}{1+\tan{\left(\arctan\frac{c_y}{c_x-1}\right)} \tan{\left(\arcsin\frac{r}{d’}\right)}}\\ &= \frac{\frac{c_y}{c_x-1}-\frac{(r/d’)}{\sqrt{1-(r/d’)^2}}}{1+\frac{c_y}{c_x-1}\cdot \frac{(r/d’)}{\sqrt{1-(r/d’)^2}}}\\ &= \frac{\frac{c_y}{c_x-1}-\frac{(r)}{k}}{1+\frac{rc_y}{k\left(c_x-1\right)}}\\ &= \frac{\frac{kc_y}{k\left(c_x-1\right)}-\frac{(r\left(c_x-1\right))}{k\left(c_x-1\right)}}{1+\frac{rc_y}{k\left(c_x-1\right)}}\\ &= \frac{kc_y-r\left(c_x-1\right)}{k\left(c_x-1\right)+rc_y}\\ \end{align*}$$

Where \(k’=\sqrt{d’^2-r^2}\). The full equation defining the boundary of where \(c\) allows \(Z_4\) to be located under \(t_3\) is

$$r_4 \left(\frac{kc_y-rc_x}{d^2}\cdot\cos{\zeta_4}+\frac{kc_x+rc_y}{d^2}\cdot\sin{\zeta_4}\right)\leq \frac{kc_y-r\left(c_x-1\right)}{k\left(c_x-1\right)+rc_y}(r_4 \left(\frac{kc_x+rc_y}{d^2}\cdot\cos{\zeta_4} - \frac{kc_y-rc_x}{d^2}\cdot\sin{\zeta_4}\right)-1)$$

We may also note that \(\cos{\zeta_4} = \frac{1-h}{r_4}\) and \(\sin{\zeta_4} = \frac{h}{r_4}\). Inputing this into the previous equation and multiplying both sides by \(d^2\) gives

$$\left(kc_y-rc_x\right)(1-h)+\left(kc_x+rc_y\right) h \leq \frac{kc_y-r\left(c_x-1\right)}{k\left(c_x-1\right)+rc_y}( \left(kc_x+rc_y\right)\cdot(1-h) - \left(kc_y-rc_x\right) h-d^2)$$

Requirements with polar coordinates:

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The equation for \(t_3\) in polar coordinates is:

$$r_{t_3}=p\csc\left(\theta-\theta_c’+\theta_d’\right)$$

Where \(p\) is the distance of this line from the origin. By plugging in the known point of \((1,0)\), this is found to be:

$$r_{t_3}=\sin\left(-\theta_c’+\theta_d’\right)\csc\left(\theta-\theta_c’+\theta_d’\right)$$

In order for \(P_1\) to lie between the origin and \(t_3\), we’ll plug the coordinates for \(P_1\) into \(r_3\leq r_{t_3}\), as demonstrated:

$$\begin{align*} r_3 &\leq \sin\left(-\theta_c’+\theta_d’\right)\csc\left(\theta_c-\theta_d+\zeta_4-\theta_c’+\theta_d’\right) \\ \sin\left(\theta_c-\theta_d+\zeta_4-\theta_c’+\theta_d’\right) &\leq \frac{\sin\left(-\theta_c’+\theta_d’\right)}{r_3} \\ \sin\left(\theta_c-\theta_d-\theta_c’+\theta_d’\right)\cos(\zeta_4) &\leq \frac{\sin\left(-\theta_c’+\theta_d’\right)}{r_3} \\ \end{align*}$$

Identifying possible locations of \(c\)

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Regardless of which formula is used, the area that allows for \(P_1\) to lie under \(t_3\) when \(h=0.2\) is shown below in blue:

By replacing \(\zeta_4\) with \(\zeta_3-\zeta_4\) and mirroring over \(y=0.5\), we get the area that allows for \(P_2\) to lie under \(t_1\). This is shown below in blue for \(h=0.2\):

Taken together, the possible locations of \(c\) is reduced substantially! The only possible region is this small trangle:

How can we define the boundary of this triangle? The upper edge of this triangle is the line \(y=r\). If we let \(y=r\), the other terms we’ve discussed simplify dramatically:

$$k=\sqrt{d^2-r^2}=\sqrt{c_x^2+r^2-r^2}=c_x$$ $$k’=1-c_x$$ $$\theta_c=\theta_d$$ $$\theta_c’=\pi-\theta_d’$$ $$P_1=\left((1-h), h\right)$$

We are curious what value of \(c_x\) causes \(P_1\) to lie under \(t_3\). The equation of \(t_3\) can be simplified to:

$$\begin{align*} y &= \tan\left(\theta_c’-\theta_d’\right)\left(x-1\right)\\ &= \tan\left(\pi-2\theta_d’\right)\left(x-1\right)\\ &= \tan\left(2\theta_d’\right)\left(1-x\right)\\ &= \frac{2\tan\theta_d’}{1-\tan^2\theta_d’}\left(1-x\right)\\ &= \frac{2rk’}{k’^2-r^2}\left(1-x\right)\\ \end{align*}$$

Plugging \(P_1\) into this gives:

$$\begin{align*} h &= \frac{2rk’}{k’^2-r^2}\left(1-(1-h)\right)\\ h &= \frac{2rk’}{k’^2-r^2}\\ k’^2-r^2 &= 2rk’\\ k’^2-2rk’-r^2 &= 0\\ \end{align*}$$

This can be solved with the quadratic formula:

$$\begin{align*} k’ &= \frac{2r\pm\sqrt{4r^2+4r^2}}{2} = \frac{2r\pm 2r\sqrt{2}}{2}\\ 1-c_x &= r(1 \pm \sqrt{2})\\ c_x &= 1-r(1 \pm \sqrt{2})\\ \end{align*}$$

By analysis, we can see that we’ll want to use the positive version of this. That gives us that \(c_x\leq 1-r(1+\sqrt{2})\). Next, let’s find the location of \(P_2\) when \(c_y=r\). This can be found with

$$\begin{align*} P_2 &= \left(1+r_4\cos\left(\pi+\zeta_4-\zeta_3\right),r_4\sin\left(\pi+\zeta_4-\zeta_3\right)\right)\\ &= \left(1-r_4\cos\left(\zeta_4-\zeta_3\right),-r_4\sin\left(\zeta_4-\zeta_3\right)\right)\\ \end{align*}$$

Just as before, we want to know when this lies under \(t_1\). But first, we’ll rewrite \(t_1\) under the assumption that \(c_y=r\):

$$y = \tan\left(\theta_c+\theta_d\right)x = \tan\left(\theta_c+\theta_d\right)x = \tan\left(2\theta_c\right)x = \frac{2\tan\theta_c}{1-\tan^2\theta_c}x= \frac{2\frac{r}{c_x}}{1-\frac{r^2}{c_x^2}}x = \frac{2rc_x}{c_x^2-r^2}x$$

And again, we plug \(P_2\) into this simplified form of \(t_1\)

$$\begin{align*} -r_4\sin\left(\zeta_4-\zeta_3\right) &= \frac{2rc_x}{c_x^2-r^2}\left(1-r_4\cos\left(\zeta_4-\zeta_3\right)\right)\\ \frac{r_4\sin\left(\zeta_4-\zeta_3\right)}{r_4\cos\left(\zeta_4-\zeta_3\right)-1} &= \frac{2rc_x}{c_x^2-r^2}\\ \end{align*}$$

In order to keep this relatively simple, I’ll define \(u=\frac{r_4\sin\left(\zeta_4-\zeta_3\right)}{r_4\cos\left(\zeta_4-\zeta_3\right)-1}\), so that this equation may be rewritten as:

$$\begin{align*} u &= \frac{2rc_x}{c_x^2-r^2}\\ uc_x^2-ur^2 &= 2rc_x\\ c_x^2-\frac{2r}{u}c_x-r^2 &= 0\\ \end{align*}$$

As before, we solve this with the quadratic formula.

$$\begin{align*} c_x &= \frac{\frac{2r}{u}\pm\sqrt{\frac{4r^2}{u^2}+4r^2}}{2}\\ c_x &= \frac{r}{u}\pm r\sqrt{\frac{1}{u^2}+1}\\ \end{align*}$$

Again, analysis lets us know to use the positive version of this. All in all, we now know that

$$c_x\in\left[\frac{r}{u}+ r\sqrt{\frac{1}{u^2}+1},1-r(1+\sqrt{2})\right]$$

Locating a third point

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I had mentioned previously that I was going to define a third point, in the same spirit as the previous two. For this, I’ll use an orientation of \(z_h\) where \(Z_1\) is located at \(\left(1,0\right)\), where \(Z_4\) lies on \(t_1\). Then, let’s define \(P_3\) to be \(Z_3\) in this position, so our three points are:

1. \(P_1=Z_4\) when \(Z_1\) is located at the origin, \(Z_2\) lies on \(t_2\)
2. \(P_2=Z_4\) when \(Z_1\) is located at \((1,0)\), \(Z_3\) lies on \(t_4\)
3. \(P_3=Z_3\) when \(Z_1\) is located at \((1,0)\), \(Z_4\) lies on \(t_1\)

Requirements for \(P_3\)

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I don’t want to spend too long discussing the requirements for \(P_3\) for a few reasons: I’m excited to get to the unique insights of this article that I’ve been preparing for, and the already existing restrictions on \(c\) and \(h\) guarantee for \(P_3\) to exist. Even so, it’s worth mentioning the assumptions necesary.

1. \(t_1\) must be close enough to \((1,0)\) for \(Z_4\) to reach it
2. \(Z_2\) has to lie inside \([m,g]\)

The second of these two requirements may be ignored. Under the position of \(z_h\) that creates \(P_3\), the angle \(\alpha\) is greater than it was for the position of \(z_h\) that creates \(P_2\). This means that \(Z_2\) will be lower than it was when we were locating \(P_2\), and thus fits into \([m,g]\) more easily. The first requirement requires more analysis. Let \(T_1\) be the point on \(t_1\) that lies on the boundary of \(m\), so that in polar coordinates:

$$T_1=\left(k,\theta_c+\theta_d\right)$$

or, in cartesian coordinates,

$$T_1=\left(k\cos\left(\theta_c+\theta_d\right),k\sin\left(\theta_c+\theta_d\right)\right)$$

We know that \Z_4\ can reach \(t_1\) when the distance of this point to \((1,0)\) is less than \(r_4\). Expressed as an equation, this becomes:

$$\begin{align*} r_4^2 &\geq k^2\sin^2\left(\theta_c+\theta_d\right) + \left(1-k\cos\left(\theta_c+\theta_d\right)\right)^2\\ r_4^2 &\geq k^2\sin^2\left(\theta_c+\theta_d\right) + 1-2k\cos\left(\theta_c+\theta_d\right)+ k^2\cos^2\left(\theta_c+\theta_d\right)\\ r_4^2 &\geq d^2 - r^2 + 1-2k\cos\left(\theta_c+\theta_d\right)\\ \end{align*}$$

Shown in green is the region of potential locations of \(c\) that satisfy this inequality. It comfortably covers the entire black region for any value of \(h\) under \(0.28\).

Again, there are ways of adjusting \(P_1\), \(P_2\), and \(P_3\) that allow variations of them to exist for larger values of \(h\). Perhaps a future version of this project will acknowledge them more specifically, but I’d like to finish the limited version of this proof and demonstrate the idea as soon as possible.

Locating \(P_3\)

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Placing \(Z_1\) at \((1,0)\) and \(Z_2\) at an angle of \(\alpha\) means that \(Z_4\) is at an angle of \(\alpha+\zeta_4) from \((1,0)\). If \(u\) is the distance from \(Z_4\) to the origin, this can be illustrated with the triangle below:

In order to understand this triangle better, we’ll use the Sine law:

$$\sin{\left(\pi-\left(\theta_c+\theta_d+\pi-\alpha-\zeta_4\right)\right)} = \frac{\sin\left(\theta_c+\theta_d\right)}{r_4}$$

or

$$\sin{\left(\pi-\alpha-\zeta_4+\theta_c+\theta_d\right)} = \frac{\sin\left(\theta_c+\theta_d\right)}{r_4}$$

Using our previously solved value of \(\sin\left(\theta_c+\theta_d\right)\), this can be rewritten in terms of \(\alpha+\zeta_4\). Just like before, it might be useful to evaluate \(\sin\left(\theta_c+\theta_d\right)\). Thankfully, the calculations are identical.

$$\begin{align*} \cos\left(\theta_c+\theta_d\right) &= \frac{kc_x-rc_y}{d^2}\\ \sin\left(\theta_c+\theta_d\right) &= \frac{rc_x+kc_y}{d^2}\\ \end{align*}$$

With this substitution, we’ll solve the last equation for \(\alpha+\zeta_4\).

$$\alpha+\zeta_4 = \pi-\arcsin\left(\frac{rc_x+kc_y}{r_4d^2}\right)+\theta_c+\theta_d$$

or

$$\alpha = \pi-\arcsin\left(\frac{rc_x+kc_y}{r_4d^2}\right)+\theta_c+\theta_d-\zeta_4$$

In polar coordinates from \((1,0)\). If we let \(\beta=\pi-\arcsin\left(\frac{rc_x+kc_y}{r_4d^2}\right)+\theta_c+\theta_d-\zeta_4+\zeta_3\), then the position of \(P_3\) can be expressed as:

$$P_3=\left(r_3, \beta\right)$$

But of course, we want this point expressed in Cartesian coordinates. The Cartesian location of \(P_3\) is given by:

$$P_3=\left(1-r_3\cos\left(\arcsin\left(\frac{\sin\left(\theta_c+\theta_d\right)}{r_4}\right)-\theta_c-\theta_d+\zeta_4-\zeta_3\right),r_3\sin\left(\arcsin\left(\frac{\sin\left(\theta_c+\theta_d\right)}{r_4}\right)-\theta_c-\theta_d+\zeta_4-\zeta_3\right)\right)$$

The distance of this point to the origin is:

$$\sqrt{\left(1+r_3\cos\beta\right)^2+\left(r_3\sin\beta\right)^2} = \sqrt{1+2r_3\cos\beta+r_3^2}$$

If we let \(\gamma\) be the angle from the origin to this point, we have:

$$\begin{align*} \frac{\sin\gamma}{r_3} &= \frac{\sin(\beta-\pi)}{\sqrt{1+2r_3\cos\beta+r_3^2}}\\ \gamma &= \arcsin\left(\frac{-r_3\sin(\beta)}{\sqrt{1+2r_3\cos\beta+r_3^2}}\right)\\ \end{align*}$$

Restricting the area where an aditional point may be placed

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In total, the three points defined so far are:

\(P_1\):

$$ \begin{array}{| c | c |} \hline \text{Location in cartesian} & \left((1-h)\frac{kc_x+rc_y}{d^2} - h\frac{kc_y-rc_x}{d^2},(1-h)\frac{kc_y-rc_x}{d^2}+h\frac{kc_x+rc_y}{d^2}\right) \\ \hline \text{Location in polar} & \left(r_4,\theta_c-\theta_d+\zeta_4\right) \\ \hline \end{array} $$

\(P_2\):

$$ \begin{array}{| c | c |} \hline \text{Location in cartesian} & \left(1+r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right),r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)\right) \\ \hline \text{Location in polar from origin} & \left(\sqrt{r_4^2+2r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)+1},\arctan\left(\frac{r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)}{r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)+1}\right)\right) \\ \hline \text{Location in polar from }(1,0) & \left(r_4, \theta_c’+\theta_d’+\zeta_4-\zeta_3\right) \\ \hline \end{array} $$

\(P_3\):

$$ \begin{array}{| c | c |} \hline \text{Location in cartesian} & \left(1+r_3\cos\beta, r_3\sin\beta\right) \\ \hline \text{Location in polar from origin} & \left(\sqrt{1+2r_3\cos\beta+r_3^2},\arcsin\left(\frac{-r_3\sin(\beta)}{\sqrt{1+2r_3\cos\beta+r_3^2}}\right)\right) \\ \hline \text{Location in polar from }(1,0) & \left(r_3, \beta\right) \\ \hline \end{array} $$

These three points are placed on the graph in green:

We also know that it is not possible to have a distance between any two points greater than 1. I’ll show in green the areas that break this rule.

This is where things get interesting. Let’s suppose we were to ask if \([m,g,P]\) accommodates \(z_h\) for any point \(P\) inside this not-green area. Not only can \(P\) not equal \(P_1\), \(P_2\), or \(P_3\), but we require that \(P_1\), \(P_2\), and \(P_3\) lie outside \([m,g,P]\). Therefore, there cannot be a point \(Q\) inside \([m,g,P]\) such that \(\langle Q,P\rangle\) contains \(P_1\), \(P_2\), or \(P_3\). It’s fairly easy to construct the equations that pass through these points and \((0,0)\) or \((1,0)\). These are given here:

$$\begin{align*} y_1 &= \frac{r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)}{1+r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)}x\\ y_2 &= \frac{r_3\sin\beta}{1+r_3\cos\beta}x\\ y_3 &= \frac{r_3\sin\beta}{r_3\cos\beta}\left(x-1\right)=\tan\beta(x-1\right)\\ y_4 &= \frac{r_4\sin\left(\theta_c-\theta_d+\zeta_4\right)}{r_4\cos\left(\theta_c-\theta_d+\zeta_4\right)-1}\left(x-1\right) \end{align*}$$

Finding the lines that connect \(P_1\) and \(P_2\) to \(m\) is a bit more involved. Finding the lines tangent to \(m\) passing through \(P_1\) and \(P_2\) would be cumbersome, and not likely to be worth the effort. Instead, let’s find the line from \(P_2\) to the end point of \(t_1\). The slope of this line is:

$$\frac{r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)-k\sin\left(\theta_d+\theta_c\right)}{1+r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)-k\cos\left(\theta_d+\theta_c\right)}$$

This gives us the line:

$$y_5-k\sin\left(\theta_c+\theta_d\right)=\frac{r_4\sin\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)-k\sin\left(\theta_c+\theta_d\right)}{1+r_4\cos\left(\theta_c’+\theta_d’+\zeta_4-\zeta_3\right)-k\cos\left(\theta_c+\theta_d\right)}\left(x-k\cos\left(\theta_c+\theta_d\right)\right)$$

Using the same method, we can find the line connecting \(P_1\) to the top of \(t_3\). This is

$$y_6-r_4\sin\left(\theta_c-\theta_d+\zeta_4\right) = \frac{k’\sin\left(\theta_c’-\theta_d’\right)-r_4\sin\left(\theta_c-\theta_d+\zeta_4\right)}{1+k’\cos\left(\theta_c’-\theta_d’\right)-r_4\cos\left(\theta_c-\theta_d+\zeta_4\right)}\left(x-r_4\cos\left(\theta_c-\theta_d+\zeta_4\right)\right)$$

Finally, using these lines as boundary points, we can make a considerably restricted area:

There are a few remarkable things about this. I’ll note that this area is incredibly tight, meaning that adding almost any point outside \([m,g]\) will allow for \(z_h\) to be accommodated. Importantly, we can conclude that if any type 1 shape does not fit inside this region, then there exists a set of three type 1 curves that accomodate \(Z_h\). Remember from previously that:

This is named as such because it is shaped as a rectangle with a side of length \(s\). Critically, there exists values of \(s\) for which \(R_s\) cannot fit inside the non-green area! In order to prove this, I’ll start by finding the intersection of \(y_1\) and \(y_4\). For convenience, I’ll define a few angles:

$$\begin{align*} \delta’ &= \theta_c’+\theta_d’+\zeta_4-\zeta_3\\ \delta &= \theta_c-\theta_d+\zeta_4\\\ \end{align*}$$

These give:

$$\begin{align*} y_1 &= \frac{r_4\sin\left(\delta’\right)}{1+r_4\cos\left(\delta’\right)}x\\ y_2 &= \frac{r_3\sin\beta}{1+r_3\cos\beta}x\\ y_3 &= \tan\beta\left(x-1\right)\\ y_4 &= \frac{r_4\sin\left(\delta\right)}{r_4\cos\left(\delta\right)-1}\left(x-1\right) \end{align*}$$

Next, let’s set \(y_1\) equal to \(y_4\) and solve for \(x\).

$$\begin{align*} \frac{r_4\sin\delta’}{1+r_4\cos\delta’}x &= \frac{r_4\sin\delta}{r_4\cos\delta-1}\left(x-1\right)\\ \frac{r_4\sin\delta’}{1+r_4\cos\delta’}x &= \frac{r_4\sin\delta}{r_4\cos\delta-1}x-\frac{r_4\sin\delta}{r_4\cos\delta-1}\\ \left(\frac{r_4\sin\delta’}{1+r_4\cos\delta’}-\frac{r_4\sin\delta}{r_4\cos\delta-1}\right)x &= -\frac{r_4\sin\delta}{r_4\cos\delta-1}\\ \left(\frac{\sin\delta}{r_4\cos\delta-1}-\frac{\sin\delta’}{1+r_4\cos\delta’}\right)x &= \frac{\sin\delta}{r_4\cos\delta-1}\\ \left(\frac{\sin\delta\left(1+r_4\cos\delta’\right)}{\left(r_4\cos\delta-1\right)\left(1+r_4\cos\delta’\right)}-\frac{\sin\delta’\left(r_4\cos\delta-1\right)}{\left(r_4\cos\delta-1\right)\left(1+r_4\cos\delta’\right)}\right)x &= \frac{\sin\delta}{r_4\cos\delta-1}\\ \frac{\sin\delta+r_4\sin\delta\cos\delta’-r_4\sin\delta’\cos\delta+\sin\delta’}{\left(r_4\cos\delta-1\right)\left(1+r_4\cos\delta’\right)}x &= \frac{\sin\delta}{r_4\cos\delta-1}\\ \frac{\sin\delta+\sin\delta’+r_4\sin\left(\delta-\delta’\right)}{\left(1+r_4\cos\delta’\right)}x &= \sin\delta\\ x &= \frac{\sin\delta\left(1+r_4\cos\delta’\right)}{\sin\delta+\sin\delta’+r_4\sin\left(\delta-\delta’\right)}\\ \end{align*}$$

Next, we can find the \(y\)-coordinate by plugging \(x\) into the equation for \(y_1\):

$$\begin{align*} y &= \frac{r_4\sin\delta’}{1+r_4\cos\delta’}x\\ &= \frac{r_4\sin\delta’}{1+r_4\cos\delta’}\cdot\frac{\sin\delta\left(1+r_4\cos\delta’\right)}{\sin\delta+\sin\delta’+r_4\sin\left(\delta-\delta’\right)}\\ &= \frac{r_4\sin\delta\sin\delta’}{\sin\delta+\sin\delta’+r_4\sin\left(\delta-\delta’\right)}\\ \end{align*}$$

I’ll define this point to be \(P_4\), so that

$$P_4=\left(\frac{\sin\delta\left(1+r_4\cos\delta’\right)}{\sin\delta+\sin\delta’+r_4\sin\left(\delta-\delta’\right)}, \frac{r_4\sin\delta\sin\delta’}{\sin\delta+\sin\delta’+r_4\sin\left(\delta-\delta’\right)}\right)$$

From here, I’m going to consider 4 different positions of \(r_s\). I’ll show that none of these can fit in the green area for some values of \(h\) and then that this suffices to show that \(r_s\) cannot fit inside the green region.

Position 1

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Let’s suppose one corner of \(r_s\) lies on \(P_4\), and that the long side of \(r_s\) lies on \(y_4\). To find the location of the opposite point on \(r_s\), let’s define \(\overrightharpoon{v_1}\) to be a vector along \(y_4\). We now have:

$$\overrightharpoon{v_1}=\left\langle 1-r_4\cos\delta, -r_4\sin\delta,\right\rangle$$

We can normalize this vector as well.

$$\hat{v_1}=\left\langle \frac{1-r_4\cos\delta}{\sqrt{1-2r_4\cos\delta+r_4^2}}, \frac{-r_4\sin\delta}{\sqrt{1-2r_4\cos\delta+r_4^2}}\right\rangle$$

The vector orthogonal to this is given by \(\left\langle \hat{v_1}[y],-\hat{v_1}[x]\right\rangle\), so the point at the bottom of \(r_s\) is given by:

$$R_{s3}=\left\langle P_4[x]+(1-2s)\hat{v_1}[x]+s\hat{v_1}[y], P_4[y]+(1-2s)\hat{v_1}[y]-s\hat{v_1}[x]\right\rangle$$

When does this point lie under \(y_2\)? We’ll plug this point into that equation:

$$P_4[y]+(1-2s)\hat{v_1}[y]-s\hat{v_1}[x] < \frac{r_3\sin\beta}{1+r_3\cos\beta}\left(P_4[x]+(1-2s)\hat{v_1}[x]+s\hat{v_1}[y]\right)$$

Solving this for \(s\) gives our first bound on \(s\):

$$\begin{align*} P_4[y]+\hat{v_1}[y]-2s\hat{v_1}[y]-s\hat{v_1}[x] &< \frac{r_3\sin\beta}{1+r_3\cos\beta}\left(P_4[x]+\hat{v_1}[x]-2s\hat{v_1}[x]+s\hat{v_1}[y]\right)\\ -2s\hat{v_1}[y]-s\hat{v_1}[x] +\frac{r_3\sin\beta}{1+r_3\cos\beta}\left(s\hat{v_1}[y]-2s\hat{v_1}[x]\right) &< \frac{r_3\sin\beta}{1+r_3\cos\beta}\left(P_4[x]+\hat{v_1}[x]\right)-P_4[y]-\hat{v_1}[y]\\ s &< \frac{P_4[y]+\hat{v_1}[y]-\frac{r_3\sin\beta}{1+r_3\cos\beta}\left(P_4[x]+\hat{v_1}[x]\right)}{2\hat{v_1}[y]+\hat{v_1}[x] +\frac{r_3\sin\beta}{1+r_3\cos\beta}\left(2\hat{v_1}[x]-\hat{v_1}[y]\right)}\\ \end{align*}$$

Position 2

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Well, what if one corner of \(r_s\) was placed at \(P_4\), but the long side of \(r_s\) lies along \(y_1\)? then let’s let the vector \(v_2\) lie along \(y_1\)

$$\overrightharpoon{v_2}=\left\langle -r_4\sin\delta’,-1-r_4\cos\delta’\right\rangle$$

As before, we can normalize this vector:

$$\hat{v_2}=\left\langle \frac{-r_4\sin\delta’}{\sqrt{1+2r_4\cos\delta’+r_4^2}}, \frac{-1-r_4\cos\delta’}{\sqrt{1+2r_4\cos\delta’+r_4^2}}\right\rangle$$

The vector orthogonal to this is given by \(\left\langle -\hat{v_2}[y],\hat{v_2}[x]\right\rangle\), so the point at the bottom of \(r_s\) is given by:

$$R_{s3}=\left\langle P_4[x]+(1-2s)\hat{v_2}[x]-s\hat{v_2}[y], P_4[y]+(1-2s)\hat{v_2}[y]+s\hat{v_2}[x]\right\rangle$$

When does this point lie under \(y_3\)? We’ll plug this point into that equation:

$$P_4[y]+(1-2s)\hat{v_2}[y]+s\hat{v_2}[x] < \tan\beta\left(P_4[x]+(1-2s)\hat{v_2}[x]-s\hat{v_2}[y]-1\right)$$

Just like before, let’s solve for \(s\):

$$\begin{align*} P_4[y]+\hat{v_2}[y]-2s\hat{v_2}[y]+s\hat{v_2}[x] &< \tan\beta\left(P_4[x]+\hat{v_2}[x]-2s\hat{v_2}[x]-s\hat{v_2}[y]-1\right)\\ -2s\hat{v_2}[y]+s\hat{v_2}[x] +\tan\beta\left(2s\hat{v_2}[x]+s\hat{v_2}[y]\right) &< \tan\beta\left(P_4[x]+\hat{v_2}[x]-1\right)-P_4[y]-\hat{v_2}[y]\\ s &< \frac{\tan\beta\left(P_4[x]+v_2[x]-1\right)-P_4[y]-v_2[y]}{\tan\beta\left(2v_2[x]-v_2[y]\right)-2v_2[y]+v_2[x]}\\ \end{align*}$$

Position 3

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Okay, what if \(r_s\) lies with a long side on \(y_2\)? Just like previously, we’ll define \(v_3\) to go along \(y_2\):

$$\overrightharpoon{v_3}=\left\langle -1-r_3\cos\beta, -r_3\sin\beta,\right\rangle$$

And normalize it:

$$\hat{v_3}=\left\langle \frac{-1-r_3\cos\beta}{\sqrt{1+2r_3\cos\beta+r_3^2}},\frac{-r_3\sin\beta}{\sqrt{1+2r_3\cos\beta+r_3^2}}\right\rangle$$

In this case, the orthogonal vector will be \(\left\langle -\hat{v_3}[y],\hat{v_3}[x]\right\rangle\), but where are we going to locate \(R_{s1}\)? First, I’ll define another point, this one being the intersection of \(y_2\) with the line \(x=1\). Using the equation for \(y_2\), we quickly get:

$$P_5=\left(1, \frac{r_3\sin\beta}{1+r_3\cos\beta}\right)$$

Then, if \(r_{s}\) lies on \(y_2\), there is some value \(a\) for which:

$$\begin{align*} R_{s1} &= \left\langle 1+a\hat{v_3}[x],P_5[y]+a\hat{v_3}[y]\right\rangle\\ R_{s2} &= \left\langle 1+a\hat{v_3}[x]+s\hat{v_3}[y],P_5[y]+a\hat{v_3}[y]-s\hat{v_3}[x]\right\rangle\\ R_{s3} &= \left\langle 1+(1-2s+a)\hat{v_3}[x]+s\hat{v_3}[y],P_5[y]+(1-2s+a)\hat{v_3}[y]-s\hat{v_3}[x]\right\rangle\\ \end{align*}$$

If we assume that \(R_{s2}\) lies on \(y_4\), we may use that to solve for \(a\):

$$\begin{align*} P_5[y]+a\hat{v_3}[y]-s\hat{v_3}[x] &= \frac{r_4\sin\left(\delta\right)}{r_4\cos\left(\delta\right)-1}\left(1+a\hat{v_3}[x]+s\hat{v_3}[y]-1\right)\\ a\hat{v_3}[y]-a\frac{r_4\sin\left(\delta\right)\hat{v_3}[x]}{r_4\cos\left(\delta\right)-1} &= s\frac{r_4\sin\left(\delta\right)\hat{v_3}[y]}{r_4\cos\left(\delta\right)-1}-P_5[y]+s\hat{v_3}[x]\\ a &= \frac{s\frac{r_4\sin\left(\delta\right)\hat{v_3}[y]}{r_4\cos\left(\delta\right)-1}+s\hat{v_3}[x]-P_5[y]}{\hat{v_3}[y]-\frac{r_4\sin\left(\delta\right)\hat{v_3}[x]}{r_4\cos\left(\delta\right)-1}}\\ \end{align*}$$

In order to save space, and give a term that doesn’t include \(s\) we’ll define \(1-2s+a=a_1+sa_2\), so that:

$$a_1=\frac{-P_5[y]}{\hat{v_3}[y]-\frac{r_4\sin\left(\delta\right)\hat{v_3}[x]}{r_4\cos\left(\delta\right)-1}}+1$$

and

$$a_2=\frac{a-a_1}{s}-2=\frac{\frac{r_4\sin\left(\delta\right)\hat{v_3}[y]}{r_4\cos\left(\delta\right)-1}+\hat{v_3}[x]}{\hat{v_3}[y]-\frac{r_4\sin\left(\delta\right)\hat{v_3}[x]}{r_4\cos\left(\delta\right)-1}}-2$$

We can use this to express \(R_{s3}\) more precicely:

$$\begin{align*} R_{s3} &= \left(1+\left(sa_2+a_1\right)\hat{v_3}[x]+s\hat{v_3}[y], P_5[y]+\left(sa_2+a_1\right)\hat{v_3}[y]-s\hat{v_3}[x]\right)\\ R_{s3} &= \left(1+a_1\hat{v_3}[x]+s\left(a_2\hat{v_3}[x]+\hat{v_3}[y]\right), P_5[y]+a_1\hat{v_3}[y]+s\left(a_2\hat{v_3}[y]-\hat{v_3}[x]\right)\right)\\ \end{align*}$$

When does this point lie above \(y_1\)? We’ll plug this point into that equation:

$$\begin{align*} P_5[y]+a_1\hat{v_3}[y]+s\left(a_2\hat{v_3}[y]-\hat{v_3}[x]\right) &> \frac{r_4\sin\left(\delta’\right)}{1+r_4\cos\left(\delta’\right)}\left(1+a_1\hat{v_3}[x]+s\left(a_2\hat{v_3}[x]+\hat{v_3}[y]\right)\right)\\ s\left(a_2\hat{v_3}[y]-\hat{v_3}[x]\right) - s\frac{r_4\sin\left(\delta’\right)}{1+r_4\cos\left(\delta’\right)}\left(a_2\hat{v_3}[x]+\hat{v_3}[y]\right) &> \frac{r_4\sin\left(\delta’\right)}{1+r_4\cos\left(\delta’\right)}\left(1+a_1\hat{v_3}[x]\right)-P_5[y]-a_1\hat{v_3}[y]\\ s &> \frac{\frac{r_4\sin\left(\delta’\right)}{1+r_4\cos\left(\delta’\right)}\left(1+a_1\hat{v_3}[x]\right)-P_5[y]-a_1\hat{v_3}[y]}{a_2\hat{v_3}[y]-\hat{v_3}[x] - \frac{r_4\sin\left(\delta’\right)}{1+r_4\cos\left(\delta’\right)}\left(a_2\hat{v_3}[x]+\hat{v_3}[y]\right)}\\ \end{align*}$$

Position 4

#

Almost there! Let’s suppose \(r_s\) lies with a long side on \(y_3\). One more time, we’ll define \(v_4\) to go along \(y_3\):

$$\hat{v_4}=\left\langle -\cos\beta, -\sin\beta\right\rangle$$

Notably, this is already a unit vector! The orthogonal vector we’ll use is \(\left\langle -\hat{v_4}[y],\hat{v_4}[x]\right\rangle\), but we have the same problem as before. I’ll define one more point, that being the intersection of \(y_3\) and \(x=0\):

$$P_6=\left(0, -\tan\beta\right)$$

Then, if \(r_{s}\) lies on \(y_3\), there is some value \(b\) for which: