The question about the existence of a cycle of a given length in a $3$-connected planar graph all faces of which are pentagonal, and also attempts to solve it led to the following problem.
Insert into triangle a planar graph with pentagonal faces only, so that the degree of each of its vertices is not less than three.
This problem was solved here. The solution is obtained as follows. Consider a plane graph with $7$ vertices:
Each face in it is a pentagon. We inscribe a dodecahedron graph into two of the three pentagons. As the result we obtain a graph with $37$ vertices all faces of which inside triangle are pentagons. The degree of each vertex of this graph is at least $3$. Denote this graph by $Q$. We can see a plane image of graph $Q$ here and here.
My question. Is graph $Q$ minimal in the number of vertices among planar graphs in which there is a single triangular face and all other faces are pentagons and the degree of each vertex of this graph is at least $3$?
This question is posed purely out of curiosity and because I could neither prove the minimality of $Q$, nor construct a smaller graph with this property.
The answers of Parcly Taxel and student91 show that $37$ was too rough estimate for graph with specified property. The graph constructed by Parcly Taxel is symmetric and especially beautiful. But I am itching to clarify my question.
Clarifying Question. After all, what is the minimal number of vertices in planar $3$-connected graphs all faces of which are pentagons except for exactly one triangular face.
As follows from Euler's formula for planar graphs, if a graph has a single triangular face and other faces are pentagons and three vertices are of degree $4$ and others are of degree $3$, then such graph must have $25$ vertices and Parcly Taxel constructed it.