I have the following question:
So I've established that the centre of mass for $x$ is $$x=\frac{(1-\alpha^2)}{\pi(1-\alpha)} $$ and $y$ is the same. My thought to prove the centre of mass was outside was to use that it would be outside if $$x^2+y^2<\alpha^2$$ meaning (since x=y) that $2x^2=(\frac{(1-\alpha^2)}{\pi(1-\alpha)})^2<\alpha^2 $. But I can't get this to work. I feel like I'm missing something obvious..
Assuming you got $x$ right, you can simplify it as
$$x=\frac{1-\alpha^2}{\pi(1-\alpha)}=\frac{(1-\alpha)(1+\alpha)}{\pi(1-\alpha)}=\frac{1+\alpha}{\pi}$$
Now,
$$x^2+y^2=2\frac{(1+\alpha)^2}{\pi^2} < \alpha^2 $$ $$2(1+\alpha)^2<\pi^2\alpha^2$$ $$2(1+\alpha)^2-\pi^2\alpha^2<0$$ $$(2-\pi^2)\alpha^2+4\alpha+2<0$$ $$(\pi^2-2)\alpha^2-4\alpha-2>0$$
A parabola with positive coefficient at $\alpha^2$ (we use that $\pi^2>2$) is positive outside its roots, of which we'll calculate the bigger one:
$$D=4^2-4\cdot(-2)\cdot(\pi^2-2)=16+8(\pi^2-2)=8\pi^2$$ $$\alpha=\frac{4+\sqrt{D}}{2(\pi^2-2)}=\frac{4+2\sqrt{2}\pi}{2(\pi^2-2)}= \frac{2+\sqrt{2}\pi}{\pi^2-2}=\frac{\sqrt{2}(\pi+\sqrt{2})}{(\pi+\sqrt 2)(\pi - \sqrt 2)} = \frac{\sqrt{2}}{\pi-\sqrt{2}}$$
This means the former inequality ($x^2+y^2<\alpha^2$) is satisfied if $\alpha > \frac{\sqrt{2}}{\pi-\sqrt{2}}$.
The inequality is also satisfied if $\alpha$ is less then the smaller root, which equals
$$\frac{4-\sqrt{D}}{2(\pi^2-2)}=\frac{4-2\sqrt{2}\pi}{2(\pi^2-2)}= \frac{2-\sqrt{2}\pi}{\pi^2-2}=\frac{\sqrt{2}(\pi-\sqrt{2})}{(\pi+\sqrt 2)(\pi - \sqrt 2)} = \frac{\sqrt{2}}{\pi+\sqrt{2}}$$
So, for $\alpha \in \bigg[0, \frac{\sqrt{2}}{\pi+\sqrt{2}} \bigg)$ the center of mass is also outside.