How many maximum number of isosceles triangle are possible in a regular polygon of $n$ sides? 
How many maximum number of isosceles triangle are possible in a regular polygon of $n$ sides?

By separable theorem, we can say that "All regular $n$-sided polygons are separable into n congruent (identical) isosceles triangles." Is this the maximum number too?
ADDED: From mjqxxxx's comment it is clear that there are infinitely many maximum number of  isosceles triangles into which a regular n-gon can be divided.So lets rephrase our question: how many  number of isoceles triangles  can be formed from the vertices of a regular n-gon?
 A: In the case where the isosceles triangles must have vertices drawn from the regular $n$-gon, let $k$ be the number of the $n$-gon sides that are outside one of the two legs of the isosceles triangle.
For every isosceles triangle $k$ must be an even number strictly between $0$ and $n$. On the other hand, if $k$ is given, then the side lengths of the isosceles triangle also follow, and the only choice we have is which of the $n$ vertices to choose as the apex. This analysis gives $\lfloor \frac{n-1}{2}\rfloor n$ possibilities. However, if $n$ is a multiple of $3$ then we have counted each of the $n/3$ equilateral inscribed triangles tree times, and we must correct for this.
In total the number of triangles is
$$ \left\lfloor \frac{n-1}{2}\right\rfloor n - \begin{cases}(2/3)n&\text{if }3|n\\0&
\text{otherwise}\end{cases}$$
Or, expressed by case analysis modulo 6:
$$ \#\text{isosceles} = \begin{cases}
n^2/2 - (5/3) n & \text{for } n\equiv 0 \pmod 6 \\
n^2/2 - (1/2) n & \text{for } n\equiv 1, 5 \pmod 6 \\
n^2/2 - n          & \text{for } n\equiv 2, 4 \pmod 6 \\
n^2/2 - (7/6) n & \text{for } n\equiv 3 \pmod 6
\end{cases}$$
