If you have a polygon with equal sides and equal distance from center to all vertices it seems to be a regular convex polygon
EDIT I've found much easier way.
Assume that center of the polygon has coordinates (x_0,y_0) and known vertice has coordinates (x_n,y_n). Also assume that we are considering n-sided polygon.
Coordinates of i-th vertce (0<i<n) can be calculated using this formulae
x_i = x_0+R*cos(a+2*Pi*i/n)
y_i = y_0+R*sin(a+2*Pi*i/n)
where
_______________________
R = v(x_n-x_0)^2+(y_n-y_0)^2
a = acos((x_n-x_0)/R)
According to your example computations using formula above shows that
A=(4, 6)
B=(0.1339745962155614, 6.2320508075688776)
C=(1.8660254037844377, 2.7679491924311228)
You can check (e.g. using this calculator) that distances between A and B, B and C, C and A are the same and equal to 3.8729833462074166. Also yu can calculate distance between center and each vertice and see that they all will be the same.
==================
That means you can find length of the side of such polygon using this formula
a=2Rsin(Pi/n), where R is a distance between center c of your poly and its known vertice p.
_____________________________
R=v(c.x - p.x)^2 + (c.y - p.y)^2
So you will have a triangle based on center c of your poly and its first p and second s vertices. Since you know coordinates of c and p and length of all sides of this triangle (distance between c and p is R, distance between p and s is a and distance between s and c is again R) you can determine coordinates of s.
We calculate explicitly that the area of the limiting polygon is $\dfrac{4}{7}A$ where $A$ is the area of the original triangle.
Note that on the $(n+1)$-th iteration we cut off twice as many triangles as we did in the $n$-th iteration. Consider the triangles we cut off in the $(n+1)$-th iteration. Each of these has $\dfrac{1}{3}$ the height and $\dfrac{1}{3}$ the base of a triangle cut off in the $n$-th iteration. Hence the ratio of the areas of these triangles is $\dfrac{1}{9}$. Since the total area of the first triangles we cut off is $\dfrac{1}{3}A$, we have that the area of the limiting polygon is
$$A-\sum\limits_{n=0}^\infty2^n\left(\frac{1}{3}A\right)\left(\frac{1}{9}\right)^n=A-\frac{1}{3}A\frac{1}{1-\frac{2}{9}}=\frac{4}{7}A$$
Whether this actually helps us determine the shape of the limiting figure I am not so sure.
This was motivated by Hagen von Eitzen's calculation and Michael's subsequent observation.
$\\$
EDIT: Explanation of $\dfrac{1}{3}$ base and $\dfrac{1}{3}$ height statement.
Consider the polygon at a particular vertex $V$ just before making the $n$-th iteration cuts. Choose a particular edge bordering $V$ and let it have length $9x$.
Now take the $n$-th and $(n+1)$-th iteration cuts. In the figure above, the green triangle is one that is cut off in the $n$-th iteration cuts, and the red triangle is one cut off in the $(n+1)$-th iteration cuts.
Because we trisect each time, one side of the green triangle has length $3x$ and one side of the red triangle has length $x$. Call these sides the base of each triangle. Consider now the length of the altitudes corresponding to these bases, letting the altitude of the green triangle be of length $3y$. By similar triangles we find that the altitude of the red triangle has length $y$ (the similar triangles are outlined in black).
Hence the ratio of area of the red triangle to the area of the green triangle is
$$\frac{\frac{1}{2}xy}{\frac{1}{2}(3x)(3y)}=\frac{1}{9}$$
as claimed.
Fun note: we did not use the fact that the triangles in question are isosceles.
Best Answer
Suppose you have a plane $n$-gon with perimeter $L$ and maximum area.
If there are two sides whose lengths differ then there are two adjacent sides $AB$ and $BC$ with different lengths. If you replace the triangle $ABC$ with an isosceles triangle $AB'C$on base $AC$ with $|AB|+|BC| = |AB'|+|B'C|$ you will increase the area of the polygon, contradicting the area maximality.
I suspect a similar argument can prove the angles are all equal. I haven't thought that through.
Edit. In fact both my argument for the equality of the side lengths and the argument for angles is the core of the answer at this question, linked from the comments: Given a polygon of n-sides, why does the regular one (i.e. all sides equal) enclose the greatest area given a constant perimeter?
To extend the question to three dimensions requires a definition of the area of a polygon that might not be planar. You will have to clarify the question.