The Four-vertex theorem states that any simple closed plane curve has at least four vertices, i.e $\kappa'(t) = 0$ where $\kappa$ is the signed curvature function.
The proof given by Osserman here depends on the lemma: Given a compact set $K$ in $\mathbb{R}^{3}$, there exists a unique smallest circle enclosing $K$ (the circumscribed circle)
My question is: how does one actually find this circle?
If I have a simple closed plane curve, defined parametrically $\alpha : I \mapsto \mathbb{R}^{2}$ and I want to find its vertices, I first try to find the signed curvature function $$\kappa(t) = \frac{\det(\alpha', \alpha'')}{||\alpha'(t)||^3} $$
then find the points where the derivative $\kappa'(t) = 0$
This almost never goes smoothly and I end up with messy calculations. So instead, using the proof, I can find such circumscribed circle and show that it intersects $\alpha$ in $n$ points, proving that the curve has indeed $2n$ vertices.
For the case that the curve has symmetry, like here
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It could work to define a circle with center at the origin and radius $R = \sup_{t \in I}||\alpha(t)||$. Is there another way all together to find the vertices of a simple closed curve?