Dear Jura, your question is interesting, and plausibly the answer will be "NO"...

Anatolij

 Dear Yurii, I am not sure I understood your question since the number of intersection points between any 2 closed curves in R^2 (counted with multiplicities) is even. Thus your condition is automatically satisfied. 

Yours B.Shapiro

Dear  Anatolij.

It seems to me that the answer will be "Yes" but I can not prove it. 

See" Integral geometry and Mizel's problem" Yu. B. Zelinskii, M. V. Tkachuk, B. A. Klishchuk

Yuri

Dear Boris

We calculate not only transversal intersections but also tangential. Therefore, the number of intersections can be odd.

Yours Yuri

Dear Yuri and followers,

On ResearchGate, there are a few  questions  and discussion related to the notion of convexity. We will numerate some question.

Q1.   James F Peters 

University of Manitoba,

What are applications of convex sets and the notion of convexity in mathematics and science?

Q2. https://www.researchgate.net/post/Any information about the source of an elementary characterization of convex functions by Lagrange theorem.

Q3. https://www.researchgate.net/profile/Yuriy Zelinskyi/questions

Is it true, that next Jordan curve be a circle?

At first lglance it seems that  we  can connect those questions.

Let $C$ be closed Jordan curve in $\mathbb{R}^2$ and (X): for arbitrary circle $L$ from property that intersection $C \cap L$ contains $3$ points follows, that $C \cap L$ contains no less then $4$ points. Is it true, that $C$ also be a circle?

If $C$  is   closed Jordan curve in $\mathbb{R}^2$ which contains $\infty$, then we consider property

(X'): for arbitrary line $L$ from property that intersection $C \cap L$ contains $3$ points follows, that $C \cap L$ contains no less then $4$ points.

Suppose that $A$ is Mobius transformation.

If $A(C)\subset \mathbb{C}$ and $C$ has property (X), then $A(C)$ has property (X).

If there is a point $z_0 \in C$ such that $A(z_0)= \infty$ and $C$ has property (X), then

$A(C)$ has property (X').

It seems that property (X') can be used to connect Q2 and Q3.

Curve $C_1$ defined by $y=x^2$ is convex curve and every line intersects $C_1$ at most $3$ points counting $\infty$.

If $B(z)= \frac{1}{z-1}$, then $B(C_1)$ does not satisfy property (X).

Roughly speaking,   it seems that Q3 can be reformulate as :

Q4. If $C$ be closed Jordan curve in $\overline{\mathbb{C}}$    which contains $\infty$ and which is not a  line,

whether   there is a line $L$ such that $L$ intersects $C$ at exactly two finite points ?      However,  two   concepts  are not equivalent.

One can try first to consider    Q4 for smooth curves.

In general there is a closed Jordan curve $C$ in $\overline{\mathbb{C}}$ such that every line $L$ intersects $C$ at infinite number of points.

Note if a line $L$ intersects a $C^1$ curve $C$ at lest three finite points, then the Gauss map is not unique.

We used property that Mobius transformations   maps  circles into circles. 

best regards,

M. Mateljevic

Dear Yuri and followers,

The question is interesting.

At a  first glance, Q3 can be reformulate as:

Q4. If $C$ is   closed Jordan curve in $\mathbb{R}^2$ which contains $\infty$ and which is not a  line,

whether there is a line $L$ such that $L$ intersects $C$ at exactly two finite points.

One can try first to cosider Q4 for smooth curves.

In general there is a closed Jordan curve $C$ in $\overline{\mathbb{C}}$ such that every line $L$ intersects $C$ at infinite number of points.

For example, consider a curve $C_1$ define by $y=sin x$ , $x\geq 0$, and a curve $C_2$ join $0$

and $\infty$ which does not intersect $C_1$ and ''spiraling'' to $\infty$ and let $C$ be curve obtained by ''adding'' curve $C_2$ on $C_1$. One can check easily that every line $L$ intersects $C$ at infinite number of points.

At a first glance,  this heuristic argument indicates that the answer to Q3 can be  negative in general ??   and positive for convex curves.

I am on vacation and perhaps I miss something here.

A very rough note related to  the convexity and the posed questions  is enclosed.

best regards,

M. Mateljevic

Dear Yuri and followers,

The question is interesting  and it seems that it can be connected   with  other subjects.

Formulate   Q3  and Q4 as  conjectures:

C3. If a  closed Jordan   curve $C$   is not a circle,   then  there is a circle $L_0$  such that  $|C \cap L_0|=3$.

Here for a set A   by   $|A|$  we denote the number of points of set $A$.

C4.  If $C$ be closed Jordan curve in $\overline{\mathbb{C}}$      which contains $\infty$ and which is not a  line,

then    there is a line $L$ such that $L$ intersects $C$ at exactly two finite points.

At a  first glance  we can think that C3 is equivalent to  C4.

It seems that C4 is   easily to visualize then  C3. 

It seems that   in general,  

C4 is not true    and   

  C3 is not equivalent to C4.

So at a first glance  use of Mobius transformations seems to be  attractive,   but it seems that in general    we need others tools to attack    the problem.

best regards,

M. Mateljevic

Dear Miodrag

Mobius transformations maps only part of circles into lines.  But others remain circles. 

best regards,  Yuri