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Let $\gamma\colon [0,1] \to \mathbb R^2$ (continuous) be a simple closed plane curve and $\,\mathcal C$ its image. Let $x,y$ be some functions such that $\gamma(t)=(x(t),y(t))$. $\gamma$ is said differentiable if $x,y$ are differentiable, $\gamma'(0)=\gamma'(1)$ and $\gamma'(t)$ is never the zero vector.

$\gamma$ is said strictly convex if for any two points in the interior of $\,\mathcal C$, the line segment that joins them stays entirely inside the interior of $\,\mathcal C$ and for any two points in $\,\mathcal C$, the line segment that joins them is not included in $\,\mathcal C$.

Question 1. Let's assume $\gamma$ is differentiable and strictly convex. Is it true that for any line $r$ in the plane there are exactly two lines tangent to $\,\mathcal C$ and parallel to $r$?

Two points $A,B \in \mathcal C$ are opposite if the tangent lines to $\,\mathcal C$ at $A$ and at $B$ are parallel. Two opposite points are minimally diametrical if their (Euclidean) distance is the minimum among the distances of opposites points.

Question 2. Is it true that there are at least two pairs of minimally diametrical points in $\,\mathcal C$? If $A,B$ are minimally diametrical, is the line through $A,B$ perpendicular to the tangent lines at $A,B$?

I was wondering whether I can answer these questions using the extreme value theorem, as it seems one of those problems when there's a continuous function on a closed interval... But I can't see how to apply it in this case.


Answer to Edu.

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Red points are on parallel lines not perpendicular to the line through them. If we force the red points to be opposite, then (I think) we can find some points (black) near the red points with parallel tangent lines and smaller distance. This would imply that the red points are not minimally diametrical.

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Question 1

Since $\gamma$ is strictly convex the total curvature of the curve is $2\pi$, which means that the winding number of the unit tangent vector about the origin is one. Thus, the tangent to the curve is parallel to the line $r$ in two points, say $t_0$ and $t_1$, and the lines defined by the tangent vector at those points are parallel to $r$.

The uniqueness of the lines is also clear by the same argument.

Question 2

As you said, you can use the Extreme Value Theorem for the distance function which is continuous in metric spaces for a fixed point. (Here the argument might be subtle, but I think if you choose the diametrical points as the points that you fix to find the extremes values you are fine.)

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  • $\begingroup$ I don't understand how to prove that the line through two diametrical points is perpendicular to the tangent lines at those points. $\endgroup$ – OliverN Oct 19 '16 at 19:29
  • $\begingroup$ @OliverN I think you can construct a counterexample for that... Take to points and the line through them. Now, take two parallel lines passing through $A$ and $B$ respectively (say $\ell_1,\ell_2$) that are not perpendicular to the first line. In this setup, can you construct an strictly convex, simple, closed curve with $A$ and $B$ diametrical points and with $\ell_1,\ell_2$ tangent lines at the points $A$ and $B$? $\endgroup$ – Edu Oct 20 '16 at 9:10
  • $\begingroup$ I don't think I can... $\endgroup$ – OliverN Oct 21 '16 at 12:40
  • $\begingroup$ Intuitively, but it may be wrong, ` A -----------+-------- / \ \ / ----+--------------- B ` if you take some points right to $B$ and left to $A$, they could be opposite but not diametrical and their difference would be less than $AB$, so $A$ and $B$ are not diametrical with minimal distance. $\endgroup$ – OliverN Oct 21 '16 at 12:46
  • $\begingroup$ @OliverN I don't come up with a proof for your question because I think it's not true (perhaps I'm wrong, though). The idea for my counterexample is to construct a closed curve similar to an ellipse, but with the tangent at the antipodal points not perpendicular to the line through those points. I think this is doable, does it convince you as well? $\endgroup$ – Edu Oct 21 '16 at 13:16

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