How many isosceles triangles do you need to make any polygon? A while back, I had a question regarding constructing shapes with only isosceles triangles. I decided to give it a go again and it has once again stumped me. The question is:
How many isosceles triangles would you need to be able to construct any $n$ sided polygon?
I thought induction could work, but dealing with non convex polygons may make this difficult. I tried by finding polygons that require what I thought was the maximum amount, but this is unreliable and may not be enough to find a general expression in terms of $n$.
How would one go about solving a problem like this? Is this a well known result? Any help or guidance would be greatly appreciated!
 A: There is a two ears theorem which states for $n > 3$, any simple $n$-gon has at least two  ears. 
If one split a $n$-gon at an ear, we get a triangle and a $(n-1)$-gon. 
For any triangle $ABC$ with $BC$ being the longest side.


*

*If $ABC$ is a right triangle, the circumcenter $O$ coincides with midpoint of $BC$, we can decompose $ABC$ into two isosceles triangle $ABO$, $AOC$.

*Otherwise, let $D$ be the foot on $BC$. We can split $ABC$ first into two 
right triangles $ABD$, $ADC$ and then into $4$ isosceles triangles. 
In general, we can decompose any triangle into at most $4$ isosceles triangles.
By induction on $n$, we find we can split a $n$-gon into at most $4n$ isosceles triangles. This bound is probably not optimal but at least we know the decomposition is always possible.
Update
For general triangle, the bound $4$ is optimal.
If triangle $ABC$ is acute, we can decompose it into $3$ isosceles triangles: $AOB$, $BOC$ and $COA$. If $ABC$ is a right triangle, $2$ is enough. This leaves us with the case of obtuse scalene triangles.
A literature search indicate in $2004$, Kosztolányi, et al${}^{\color{blue}{[1]}}$ has studied the problem of decomposing obtuse scalene triangles into $3$ isosceles triangles.
With help of a computer, they found there are $23$ families of solutions. 
Let $\alpha > 90^\circ > \beta > \gamma$ be the angles of the trangle.
In all these solutions, $\alpha, \beta$ are rational linear combinations of $180^\circ$ and $\gamma$.
In particular, this implies triangle with angles
$$(\alpha,\beta,\gamma) = \left( (5-\sqrt{2})\cdot 30^\circ, 30\sqrt{2}^\circ, 30^\circ \right)$$
cannot be decomposed into $3$ isosceles triangles. It is not hard to verify we cannot decompose this triangle into $2$ isosceles triangles. This means for general triangle, the bound $4$ is optimal.
Notes


*

*$\color{blue}{[1]}$ - Kosztolányi, József & Kovács, Zoltán & Nagy, Erzsébet. (2004). Decomposition of triangles into isosceles triangles II. Complete solution of the problem by using a computer. Teaching Mathematics and Computer Science. 2. 275-300. 10.5485/TMCS.2004.0059. 
An online copy can be found here.
A: If I combine the isosceles triangle side by side, I'll create a different polygon, but at some certain combination, it will still produce same shape
Just like when I join two parallelogram side by side, it also creates a parallelogram which is still a quadrilateral
so I'll assume your question about the minimum number of isosceles triangle needed to make a polygon of $n$ sides
$1$ ∆ is a $3$ sided polygon
$2$ ∆ makes a quadrilateral 
$3$ ∆ makes a 5 sides polygon
so at least $n$ number of isosceles triangle makes a $(n+2)$ sided polygon
