A line perpendicular to a tangent curved at both ends

Question

I made a guess two years ago. I have a strong feeling that there is a proof using the fixed point theorem with geometric visualization, but I couldn't do the proof.

If you have a simple closed convex curve, thr line Perpendicular to the tangent on the curve at both ends must be present

If this feature is already known, please point to a reference that mentions it

If you can prove it would be appreciated, thanks

Answer 1

As I understood your question, the following fact (which should be well-known) provides a positive answer to it. Let $I$ be the unit segment $[0,1]$ endowed with the usual topology. Let $\varphi:I\to \mathbb R^2$ be a closed curve, that is a continuous map such that $\varphi(0)=\varphi(1)$. For each $(t_1,t_2)\in I^2$ put $f(t_1,t_2)$ equal to the inner product $(\varphi(t_1)-\varphi(t_2),\varphi(t_1)-\varphi(t_2))$, that is to the squared distance between the points $\varphi(t_1)$ and $\varphi(t_2)$. Then the function $f$ is continuous, being a composition of continuous functions. Since the set $I^2$ is compact, the function $f$ attains its maximum at some point $(s_1,s_2)\in I^2$. Suppose that for some $i\in\{1,2\}$ the function $\varphi$ is differentiable at the point $s_i$ and $\varphi'(s_i)\ne 0$ (in the case $s_i=0$ or $s_i=1$ we additionally require that $\varphi'(0)=\varphi'(1)$). Then $$0=\frac{\partial f}{\partial t_i}{\huge|}_{s_i}=2(\varphi'(s_i),\varphi(s_i))-2(\varphi'(s_i),\varphi(s_{3-i}))= 2(\varphi'(s_i),\varphi(s_i)-\varphi(s_{3-i})),$$ which means that the vector $\varphi'(s_i)$ is perpendicular to the segment $[\varphi(s_1),\varphi(s_2)]$.

Answer 2

The convexity assumption can be removed. One only need to have a simple closed differentiable curve.

Call $\gamma : [a,b] \to \mathbb{R}^2$ the curve. Assume that $\gamma$ is differentiable, that its derivative does not vanish anywhere, that $\gamma(a)=\gamma(b)$ and $\gamma'(a)=\gamma'(b)$. Hence $\gamma$ can be extended into a $b-a$ - periodic differentiable map from $\mathbb{R}$ to $\mathbb{R}^2$.

The diameter of $\gamma([a,b]) := \sup\{|\gamma(t_2)-\gamma(t_1)| : (t_1,t_2) \in \mathbb{R}^2\}$ is not $0$ and achieved at some couple $(t_1,t_2)$. Therefore, $$2 \langle \gamma(t_2)-\gamma(t_1),\gamma'(t_2) \rangle = \frac{d}{dt}\big(|\gamma(t)-\gamma(t_1)|^2\big)\Big|_{t=t_2} = 0,$$ so $\gamma'(t_2)$ is orthogonal to $\gamma(t_2)-\gamma(t_1)$. In the same way $\gamma'(t_1)$ is orthogonal to $\gamma(t_2)-\gamma(t_1)$.