= Solution
Here the <surface> is the <real projective plane>, with <Euler characteristic> one. The bound gives $V\geq6$. The following ten triangles on vertices $1,\ldots,6$ attain it:
$$
\boxed{123,\ 134,\ 145,\ 156,\ 126,\ 235,\ 245,\ 246,\ 346,\ 356.}
$$
Take these as the maximal simplices of an abstract <simplicial complex>, identifying edges with the same vertex labels. Every one of the 15 pairs of vertices occurs as an edge in exactly two triangles. The links of the six vertices are single five-cycles; listed cyclically they are
$$
\begin{array}{c|l}
1&2,3,4,5,6\\
2&1,3,5,4,6\\
3&1,2,5,6,4\\
4&1,3,6,2,5\\
5&1,4,2,3,6\\
6&1,5,3,4,2
\end{array}
$$
Thus every point has a disk neighborhood, and the realization is a connected closed <surface>. Its <Euler characteristic> is $6-15+10=1$, so the <classification of compact surfaces> identifies it with the <real projective plane>. This is an explicit <triangulation>, proving $\boxed{V_{\min}=6}$.
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