A convex polygon $P$ having an interior point $A$ in generic position (not on any line defined by two vertices) can be encoded combinatorially by associating to every vertex $v$ of $P$ the unique edge $e(v)$ of $P$ such that $A$ is contained in the triangle defined as the convex hull of $v$ and $e(v)$. (Equivalently, $e(v)$ is the unique edge whose interior intersects the line containing the two points $v$ and $A$.)

To what extend does the converse hold: given a cyclically ordered set of $n$ abstract "vertices" $v_1,\dots,v_n$ and a function $e$ which associates to every vertex an "edge" consisting of two other cyclically consecutive vertices, does the function $e$ arise as above as the function indicating the combinatorial position of a fixed interior point for a suitable planar realization of the points $v_1,\dots,v_n$ in convex position?

A necessary condition is of course that any two "abstract" triangles intersect, ie. the intersection of two distinct "triangles" should have strictly positive area in any geometric realization of the $n$ cyclically ordered points in convex position.

Are there other restrictions? (There are none for $n\leq 5$ at least.)

Added: The combinatorial situation depicted above can be encoded as follows: One consider $n$ red beans (vertices) $r_1,\dots,r_n$ and $n$ black beans ("edges" or more precisely the intersection of a line through a vertex and the interior point $A$) $b_1,\dots,b_n$ on a circle $S$ where the indices of both read and black beans correspond to the induced cyclic order.
We require moreover that every black bean $b_i$ is separated from the read bean $r_i$ by at least one other read bean on both intervals of $S\setminus{b_i,r_i}$. This situation gives rise to an involution since the last condition holds then also with colors reversed. In particular, if the necessary condition above is sufficient for a given $n$, then there is also always a "dual" configuration for every convex polygon together with a generic interior point.

Second PS: Zaimi's proof shows that such configurations come indeed in pairs as the read and black bean model shows. A direct proof is as follows: Given a realization with vertices corresponding to read beans, define the black vertices (beans) by intersecting all $n$ lines through $A$ and a vertex. A small perturbation into convex position of the black vertices realizes the dual configuration.

Zaimi's proof implies that we can count (up to combinatorial orientation-preserving equivalence or up to isotopy, they coincide in this case) the number of such configuration with $n$ vertices in convex position and an additional interior point. There are $$-1+\frac{1}{2n}\sum_{d\vert n,\ d\equiv 1\pmod 2}\phi(d)2^{n/d}$$ of them with $\phi(d)$ Euler's indicator function defined by $\phi(1)=1$ and $\phi(n)$ given by the number of invertible integers modulo $n$ for $n>1$. The first non-zero terms are $1,1,3,5,9,15,29,51,93,171,\dots$ for $n=3,4,\dots$.

5 added 282 characters in body

A convex polygon $P$ having an interior point $A$ in generic position (not on any line defined by two vertices) can be encoded combinatorially by associating to every vertex $v$ of $P$ the unique edge $e(v)$ of $P$ such that $A$ is contained in the triangle defined as the convex hull of $v$ and $e(v)$. (Equivalently, $e(v)$ is the unique edge whose interior intersects the line containing the two points $v$ and $A$.)

To what extend does the converse hold: given a cyclically ordered set of $n$ abstract "vertices" $v_1,\dots,v_n$ and a function $e$ which associates to every vertex an "edge" consisting of two other cyclically consecutive vertices, does the function $e$ arise as above as the function indicating the combinatorial position of a fixed interior point for a suitable planar realization of the points $v_1,\dots,v_n$ in convex position?

A necessary condition is of course that any two "abstract" triangles intersect, ie. the intersection of two distinct "triangles" should have strictly positive area in any geometric realization of the $n$ cyclically ordered points in convex position.

Are there other restrictions? (There are none for $n\leq 5$ at least.)

Added: The combinatorial situation depicted above can be encoded as follows: One consider $n$ red beans (vertices) $r_1,\dots,r_n$ and $n$ black beans ("edges" or more precisely the intersection of a line through a vertex and the interior point $A$) $b_1,\dots,b_n$ on a circle $S$ where the indices of both read and black beans correspond to the induced cyclic order.
We require moreover that every black bean $b_i$ is separated from the read bean $r_i$ by at least one other read bean on both intervals of $S\setminus{b_i,r_i}$. This situation gives rise to an involution since the last condition holds then also with colors reversed. In particular, if the necessary condition above is sufficient for a given $n$, then there is also always a "dual" configuration for every convex polygon together with a generic interior point.

Second PS: Zaimi's proof shows that such configurations come indeed in pairs as the read and black bean model shows. A direct proof is as follows: Given a realization with vertices corresponding to read beans, define the black vertices (beans) by intersecting all $n$ lines through $A$ and a vertex. A small perturbation into convex position of the black vertices realizes the dual configuration.

4 added 117 characters in body

A convex polygon $P$ having an interior point $A$ in generic position (not on any line defined by two vertices) can be encoded combinatorially by associating to every vertex $v$ of $P$ the unique edge $e(v)$ of $P$ such that $A$ is contained in the triangle defined as the convex hull of $v$ and $e(v)$. (Equivalently, $e(v)$ is the unique edge whose interior intersects the line containing the two points $v$ and $A$.)

To what extend does the converse hold: given a cyclically ordered set of $n$ abstract "vertices" $v_1,\dots,v_n$ and a function $e$ which associates to every vertex an "edge" consisting of two other cyclically consecutive vertices, does the function $e$ arise as above as the function indicating the combinatorial position of a fixed interior point for a suitable planar realization of the points $v_1,\dots,v_n$ in convex position?

A necessary condition is of course that any two "abstract" triangles intersect, ie. the intersection of two distinct "triangles" should have strictly positive area in any geometric realization of the $n$ cyclically ordered points in convex position.

Are there other restrictions? (There are none for $n\leq 5$ at least.)

Added: The combinatorial situation depicted above can be encoded as follows: One consider $n$ red beans (vertices) $r_1,\dots,r_n$ and $n$ black beans ("edges" or more precisely the intersection of a line through a vertex and the interior point $A$) $b_1,\dots,b_n$ on a circle $S$ where the indices of both read and black beans correspond to the induced cyclic order.
We require moreover that every black bean $b_i$ is separated from the read bean $r_i$ by at least one other read bean on both intervals of $S\setminus{b_i,r_i}$. This situation gives rise to an involution since the last condition holds then also with colors reversed. In particular, if the necessary condition above is sufficient for a given $n$, then there is also always a "dual" configuration for every convex polygon together with a generic interior point.

Second PS: Zaimi's proof shows that such configurations come indeed in pairs as the read and black bean model shows.

3 added 846 characters in body
2 added 115 characters in body
1