Strictly convex curves and lines

Let γ:[0,1]→R2\gamma\colon [0,1] \to \mathbb R^2 (continuous) be a simple closed plane curve and C\,\mathcal C its image. Let x,yx,y be some functions such that γ(t)=(x(t),y(t))\gamma(t)=(x(t),y(t)). γ\gamma is said differentiable if x,yx,y are differentiable, γ′(0)=γ′(1)\gamma'(0)=\gamma'(1) and γ′(t)\gamma'(t) is never the zero vector.

γ\gamma is said strictly convex if for any two points on C\,\mathcal C, the line segment that joins them stays entirely inside the interior of C\,\mathcal C (as defined by the Jordan curve theorem), except for the endpoints.

The tangent line to γ\gamma at t∈[0,1[t \in [0,1[ is the set
{γ(t)+rγ′(t):r∈R}. \{ \gamma(t)+r\gamma'(t) : r \in \mathbb R\} .

Is this true?

If γ\gamma is strictly convex, then any line crosses C\mathcal C in at most two points. If γ\gamma is also differentiable, then a line ss crosses C\mathcal C in just one point iff ss is tangent to γ\gamma. No interior point of C\mathcal C are on a tangent line.

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To answer the first part of your question, imagine a construction where a line ll crosses C\mathcal{C} at points aa, bb, and cc. WLOG, let cc be between aa and bb on ll. Let cϵc_{\epsilon} and c−ϵc_{-\epsilon} be displaced from cc by small arc lengths ϵ\epsilon and −ϵ-\epsilon along C\mathcal{C}, respectively (where c−ϵc_{-\epsilon} is between aa and cc, and cϵc_{\epsilon} is between cc and bb). What about the line segments connecting aa and c−ϵc_{-\epsilon}, c−ϵc_{-\epsilon} and cϵc_{\epsilon}, and cϵc_{\epsilon} and bb? Are they all on the interior of C\mathcal{C}?
– Michael Lee
2 days ago

  

 

Maybe I should modify my definition…
– OliverN
2 days ago

  

 

See the edit. Now I can say that if cc is between aa and bb, then it is in the interior of C\mathcal C, impossible. Is it correct?
– OliverN
2 days ago

  

 

That looks absolutely correct. The second claim should also be true by boundary arguments.
– Fimpellizieri
2 days ago

  

 

Making the second claim more precise; since γ′(t)\gamma'(t) is never zero, one can always use the implicit function theorem to write γ\gamma as the graph of a function, locally. Now, it’s easy to show that if a line touches a graph at a point and the derivatives do not agree on that point, then the line crosses the graph. Can you see how this solves the problem?
– Fimpellizieri
2 days ago

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