5. (CZS 1) (SL77-2).

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5. (CZS 1) (SL77-2).
5. Let us denote by $I$ and $E$ the sets of interior boundary points and exterior boundary points. Let $A B C D$ be the square inscribed in the circle $k$ with sides parallel to the coordinate axes. Lines $A B, B C, C D, D A$ divide the plane into 9 regions: $\mathcal{R}, \mathcal{R}_{A}, \mathcal{R}_{B}$, $\mathcal{R}_{C}, \mathcal{R}_{D}, \mathcal{R}_{A B}, \mathcal{R}_{B C}, \mathcal{R}_{C D}, \mathcal{R}_{D A}$. There is a unique pair of lattice points $A_{I} \in \mathcal{R}, A_{E} \in$ $\mathcal{R}_{A}$ that are opposite vertices of a unit square. We similarly define $B_{I}, C_{I}, D_{I}, B_{E}, C_{E}, D_{E}$. Let us form a graph $G$ by connecting each point from $E$ lying in $\mathcal{R}_{A B}$ (respectively $\mathcal{R}_{B C}, \mathcal{R}_{C D}, \mathcal{R}_{D A}$) to its upper (respectively left, lower, right) neighbor point (which clearly belongs to $I$). It is easy to see that: (i) All vertices from $I$ other than $A_{I}, B_{I}, C_{I}, D_{I}$ have degree 1. (ii) $A_{E}$ is not in $E$ if and only if $A_{I} \in I$ and $\operatorname{deg} A_{I}=2$. (iii) No other lattice points inside $\mathcal{R}_{A}$ belong to $E$.  Thus if $m$ is the number of edges of the graph $G$ and $s$ is the number of points among $A_{E}, B_{E}, C_{E}$, and $D_{E}$ that are in $E$, using (i)-(iii) we easily obtain $|E|=m+s$ and $|I|=m-(4-s)=|E|+4$.