I am confused by a step made in a proof of the following result.
Let $f_{2}^{\text{min}}(n)$ denote the maximum number of times the minimum distance can occur among n points in the plane. Then $f_{2}^{\text{min}}(n) = \lfloor 3n - \sqrt{12n-3} \rfloor$.
Proof: Assume $n \geq 3$ and consider a set $P$ of $n$ points with minimum distance 1, and connect two elements of $P$ by a segment if and only if their distance is exactly 1. Thus, we obtain a graph $G$ embedded in the plane. Assume that $G$ has the largest possible number of edges; that is, $|E(G)|=f_{2}^{\text{min}}(n)$. It is easy to see that every vertex of $G$ is adjacent to at least two other vertices. Moreover, $G$ is two-connected; that is, $G$ remains connected after the removal of any of its vertices. The outer face of $G$ is bounded by a simple closed polygon $C$. Let $b$ and $b_{d}$ denote the total number of vertices of this polygon and the number of those vertices that have degree $d$ in $G$, respsectively. Clearly, $b = b_{2} + b_{3} + b_{4} + b_{5}$. The internal angle of $C$ at a vertex of degree $d$ is at least $(d-1)\frac{\pi}{3}$, and the sum of these angles is $(b-2)\pi$. Hence, $b_{2} + 2b_{3} + 3b_{4} + 4b_{5} \leq 3b-6$. [Important Part]:
On the other hand, denoting by $f_{i} (i \geq 3)$ the number of internal faces of $G$ with $i$ sides, we obtain from Euler's polyhedral formula that, $n - f_{2}^{\text{min}}(n) + f_{3} + f_{4} + ... = 1$
The proof then continues and I understand everything before and after where it is mentioned that $n - f_{2}^{\text{min}}(n) + f_{3} + f_{4} + ... = 1$, but how is this arrived at? It is likely a very very simple answer, but if you could explain with details how this is arrived at by Euler's polyhedral formula I would be able to understand the proof. Thank you.
EDIT: I'm concerned with either my understanding that $V - E + F = 2$ for any graph embedding or convex polyhedra, or Gerry Myerson's answer that the Euler characteristic should be 1? Can anyone else comment on this?