Maximal number of faces

In a planar graph on \(n\) vertices:

Variant
Determine the maximal number of inner faces.
Solution
If a face is not triangular, it can be divided by adding additional edges to two or more faces until all faces are triangular.

According to Euler’s formula, the graph now has \( 3n-6 \) edges, from which it is possible to create \( \frac{2}3 (3n-6) \) triangular faces, one of which is the outer face.
Answer
A given planar graph can have at most \( 2n-5 \) of inner faces.
Variant
The same problem with the additional condition that the outer face of the graph is bounded by a cycle of length \( k \).
Solution
We complete any graph with the maximum number of inner faces to a triangulation by adding \( k-3 \) edges inside the outer face. This step yields \( k-3 \) extra triangular faces. The resulting graph has a total of \( 2n-4 \) faces, out of which \((2n-4)-1-(k-3)=2n-k-2\) correspond to the original inner faces.
Answer
A given planar graph can have at most \( 2n-k-2 \) of inner faces.