# 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?

• 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, 2011 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? Dec 27, 2011 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. Dec 27, 2011 at 12:52
• @Kannappan Sampath:Now, I want to know both. Dec 27, 2011 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 $$\left\lfloor \frac{n-1}{2}\right\rfloor n - \begin{cases}(2/3)n&\text{if }3|n\\0& \text{otherwise}\end{cases}$$ Or, expressed by case analysis modulo 6: $$\#\text{isosceles} = \begin{cases} n^2/2 - (5/3) n & \text{for } n\equiv 0 \pmod 6 \\ n^2/2 - (1/2) n & \text{for } n\equiv 1, 5 \pmod 6 \\ n^2/2 - n & \text{for } n\equiv 2, 4 \pmod 6 \\ n^2/2 - (7/6) n & \text{for } n\equiv 3 \pmod 6 \end{cases}$$