# 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?

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."

$4$ is enough: consider a $1 \times 3$ rectangle. Every point in the middle $1 \times 1$ square (and a couple more) satisfies your condition.
Indeed, there is no solution for a triangle $ABC$. Suppose otherwise; choose the endpoints of the longest side (WLOG, denote $AB$). Clearly, the last vertex $C$ has to be the nearest neighbor for both of these endpoints. But we need a point $P$ in the interior of the triangle that satisfies $AP + BP > AC + BC$. You should be able to make some argument as to why this is impossible (I see a calc argument with Pythagoras, but may be a little messy; there's probably an easier way to do it).