Eliminating Type 2 Curves (hopefully)#
Is the curve \(Z\) proxy accommodated by \(\lbrace L, M, Z’ \rbrace\)?
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 \(z’\)
Let \(\omega\) be a
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.
Second attempt#
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