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?

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Please clarify what are you looking for: Are you looking for the number of isoceles triangles that can be formed from the vertices of the polygon or the number of isoceles triangles into which a regular polygon can be divided? –  user21436 Dec 27 '11 at 12:30
If you number the corners of a regular pentagon from $1$ to $5$, corners $1$, $3$, and $4$ form an isosceles triangle. Are these counted? –  Ｊ. Ｍ. Dec 27 '11 at 12:34
If the question is about dissection (as the link seems to indicate), then no, there is no maximum number of congruent isosceles triangles into which a regular $n$-gon can be divided. In particular, after the initial dissection into $n$ isosceles triangles, each resulting triangle may be divided into $4$ smaller isosceles triangles by connecting the midpoints of its sides. This may be repeated indefinitely; so at least all numbers of the form $4^k n$ are feasible. –  mjqxxxx Dec 27 '11 at 12:52
@Kannappan Sampath:Now, I want to know both. –  Quixotic Dec 27 '11 at 12:53

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 $$\left\lfloor \frac{n-1}{2}\right\rfloor n - \cases{(2/3)n&\text{if }3|n\\0& \text{otherwise}}$$ Or, expressed by case analysis modulo 6: $$\#\text{isosceles} = \begin{cases} n^2/2 - (5/3) n & \text{for } n\equiv 0 \pmod 6 \\ n^2/2 - (1/2) n & \text{for } n\equiv 1, 5 \pmod 6 \\ n^2/2 - n & \text{for } n\equiv 2, 4 \pmod 6 \\ n^2/2 - (7/6) n & \text{for } n\equiv 3 \pmod 6 \end{cases}$$