What is the smallest n-gon such that there can be an interior point further from all boundary points than the points are from each of their neighbors?

Question

This is a "sequel" question to What is the maximum n-gon such that there can't be an interior point further from all boundary points than the boundary points are from each other?

In the previous question I asked:

Given a polygon with n vertices, how small can n be before it is impossible to have an interior point p such that the distance from any vertex b to p is greater than the distance from b to its nearest adjacent vertex, b'?

For example, for any triangle, it seems that an interior point will be closer to at least one vertex than that vertex is to its nearest neighbor.

From my own inspection, it seems like the answer might be 6 - a regular hexagon is the first shape for which a point at the center is the same distance from each vertex as the vertices are from their nearest neighbors.

However, I can't figure out how to formulate a rule, or apply the intuition from the regular hexagon to irregular shapes.

For my use case, it would be useful to know "below n vertices, there is no way to have such an interior point."

My updated question is: how small can $n$ be before it is impossible to have a point $p$ such that the distance from any vertex $b$ to $p$ is greater than the distance from $b$ to either of its adjacent vertices, $b'$?

Answer 1

Let $A$ and $B$ be a pair of consecutive vertices and $P$ be the point in question. If $PA$ and $PB$ are both longer than $AB$, then angle $P$ in triangle $ABP$ has to measure less than $60°$. To make up a full revolution there have to be seven or more different angles of this type at $P$, each opposite a different side of the polygon.

So there are no candidates with fewer than seven sides. But the center of a regular heptagon is farther from the vertices than the length of any side making seven sides a sharp lower bound.