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It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work of Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ has (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects $\gamma$ in more than 4 points?

Note that if $\gamma$ lies on $\mathbb{S}^{2}$, then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.

It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work of Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ is a simple closed convex curve with (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects it in more than 4 points?

Note that if $\gamma$ lies on $\mathbb{S}^{2}$, then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.

It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ has (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects $\gamma$ in more than 4 points?

Note that if $\gamma$ lies on $\mathbb{S}^{2}$, then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.

It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work of Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ has (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects $\gamma$ in more than 4 points?

Note that if $\gamma$ lies on $\mathbb{S}^{2}$, then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.

It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ has (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects $\gamma$ in more than 4 points?

Some thoughts: Suppose that $\gamma$ lies on $\mathbb{S}^{2}$. Then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.

It is well known that any smooth simple closed convex curve $\gamma$ in $\mathbb{R}^{3}$ that meets no plane in more than 4 points has exactly 4 vertices, i.e., points of vanishing torsion; here "convex" means that the curve lies on the boundary of its convex hull. This result, as mentioned for example by Pohl in On a theorem related to the four-vertex theorem, goes back to the work Scherk and Segre in the 1930s.

I was wondering about the other direction:

Question. If $\gamma$ has (nonvanishing curvature and) exactly 4 vertices, does it then follow that no plane intersects $\gamma$ in more than 4 points?

Note that if $\gamma$ lies on $\mathbb{S}^{2}$, then points of vanishing torsion correspond to critical points of the geodesic curvature. So to find a counterexample it would suffice to construct a spherical curve with exactly 4 critical points of geodesic curvature that does not satisfy the given intersection condition.