Any ideas???

What we want is the minimal number of triangles a $n$-gon could be split. So, in case this splitting is a triangulation of the $n$-gon we have exactly $n-2$ triangles. The point is, there may appear additional vertices of the triangulation that are not vertices of the original $n$-gon. Intuitively, if we add additional points to make a triangulation, we get more triangles than without those points. So, it all boils down to prove this claim. Denote by $G$ the planar graph, with vertices and edges all the vertices/edges of the triangles. Let the number of vertices of $G$ be $n$ and $t$ be the number of triangles (the original polygon is split into). Clearly $n_1\ge n.$ By Euler's formula, we have $$ v+f-e=2\qquad (1)$$where $v,f,e$ are correspondingly the number of vertices, edges and faces of $G$. We include the outer face of $G$ also as a face, hence $f=t+1$. Think of $G$ as being on a sphere and add additional vertex $v_0$ situated at some point on the outer face. Connect $v_0$ with all the vertices of the polygon and let it be the graph $G'.$ Thus, we add one additional vertex, $n$ additional edges and also $n$ additional faces. Since $G'$ is also planar we have $$v'+f'-e'=2 \qquad (2)$$Further, $v'=v+1=n_1+1, f'=f+n=t+1+n, e'=e+n.$ Note that, $G'$ is a triangulation of the plane (I'd rather say the sphere). In this case, a well know fact says $e'=3v'-6.$ (It can be easily calculated by noticing that every face corresponds to $3$ edges and every edge corresponds to $2$ faces. It means $3f'=2e'$. Now, use $(2)$). Thus, $e+n=3(n_1+1)-6$, so $e=3n_1-n-3$. Putting all this together into $(1)$ we have $$n_1 +t+1 -3n_1+n+3=2$$which yields $t=2n_1-n -2\ge n-2.$ So, indeed, the number of triangles the polygon is split into, cannot be less than $n-2$. It is attained when we just triangulate the $n$-gon, hence it's the minimal possible number.

Nice solution

The answer is $\lceil n/3\rceil$ and as an example consider $n=6$ and a splitting of a concave hexagon as follows. [asy][asy] pair A = dir(-30); pair B = dir(90); pair C = dir(210); pair D = (2*C + A) / 3; pair E = dir(-90); pair F = (C + 2*A) / 3; filldraw(A--B--C--cycle, paleblue, blue); filldraw(D--E--F--cycle, pink, red); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); [/asy][/asy] In this case, the minimum required number of triangles is clearly $2$. Generally, it's possible to construct a concave $n$-gon composed of $\lceil n/3\rceil$ triangles. Now, if $t$ is the number of triangles in a splitting, we must prove that $n\le 3t$. But clearly, each vertex of the polygon corresponds to at least one vertex of a triangle, and each triangle has $3$ vertices. This completes the proof.

Hmm, was that what the problem was about? To find a pathological poligon like that. I mean, in this case I have solved a different problem. What I showed was that there does not exist a method of triangulation (in the usual meaning, but adding additional points is allowed) that results in less than $n-2$ trianles.

dgrozev wrote: Hmm, was that what the problem was about? To find a pathological poligon like that. I mean, in this case I have solved a different problem. What I showed was that there does not exist a method of triangulation (in the usual meaning, but adding additional points is allowed) that results in less than $n-2$ trianles. Yeah, I think so. I found the official solutions here and it seems that what they refer to is different to the usual meaning of triangulation.