Task number: 3845
Consider only those drawings of planar graphs in which each face is bounded by a cycle and where different faces may have at most one edge in common. We do not consider graphs without edges.
In each such graph \( G \) we look for a suitable vertex \( u \) so that by removing it the number of faces will decrease as much as possible. What decrease can we always guarantee? If more than one option applies, select the highest one.
The smallest degree in the \( G \) graph is 3 – if it was smaller, some face is not bounded by a cycle (degree 1) or some faces have more edges in common (degree 2).
By removing the vertex of degree \( k \), the \( k \) faces are merged into one, ie the number of faces decreases by \( k-1 \).
In general, the existence of a vertex of higher degree cannot be guaranteed, see e.g. a drawing of \( K_4 \).
The correct answer is a.