Posted in MO since it has been open in MSE for over 2 months.
The area of the largest triangle that can be inscribed in a circle of raidus $1$ is $\displaystyle \frac{3 \sqrt{3}}{4}$ for a equilateral triangle and it also gives the perimeter is $3\sqrt{3}$. For any $\displaystyle 0 \le x \le \frac{3 \sqrt{3}}{4}$ we can ask the minimum perimeter $f(x) $and maximum perimeter $g(x)$ of triangles with area $x$ inscribed in a circle of radius $1$. I ran a simulation to calculate the values of $f(x)$ and $g(x)$ for all $x$ in this interval and obtained the plot below.
For the maximum perimeter, trivially $g(0) = 4$ and $\displaystyle g\left(\frac{3 \sqrt{3}}{4}\right) = 3 \sqrt{3}$. The graph of $g(x)$ appears to be a straight line as shown by the blue line. Fitting a straight line through these two points gives the model
$$ g(x) = 4 + \left(4 - \frac{16}{3\sqrt{3}}\right)x \tag 1 $$
Can this be proven? For the minimum perimeter we have trivially $f(0) = 0$ and $\displaystyle f\left(\frac{3 \sqrt{3}}{4}\right) = 3 \sqrt{3}$ and $f(1) = 2+2\sqrt{2}$ corresponding to an isosceles right triangle. Fitting a curve with the data for $g(x)$ suggests a model of the form $f(x) = (2+\sqrt{2})x^a$ with $a \approx 0.306$ however this model with inconsistent with the fact that $\displaystyle f\left(\frac{3 \sqrt{3}}{4}\right) = 3 \sqrt{3}$ hence this model was discarded.
Question: What is the minimum and the maximum perimeter of a triangle with area $x$ and circumradius $1$?
We can show that the maximum and minimum perimeters occur precisely when the triangle is isosceles.
Say the centre of the circle is $O$, and the triangle's vertices are $A$, $B$, and $C$.
Let $\theta_A=\frac12 \angle BOC$, $\theta_B=\frac12 \angle COA$ and $\theta_C=\frac12 \angle AOB$.
The area of the triangle is $$\Delta=\frac12 \left(\sin 2\theta_A+\sin 2\theta_B+\sin 2\theta_C\right)$$
and its perimeter is $$p=2 \left(\sin\theta_A+\sin\theta_B+\sin\theta_C\right)$$
Also, $\theta_A+\theta_B+\theta_C=\pi$, and each angle lies in the interval $(0,\pi)$.
We wish to find extreme values of $p$ given a fixed $\Delta$; we can use Lagrange multipliers for this. Define $$F \left(\theta_A,\theta_B,\theta_C,\mu,\nu \right)=2 \left(\sin\theta_A+\sin\theta_B+\sin\theta_C\right) + \mu \left(\sin 2\theta_A+\sin 2\theta_B+\sin 2\theta_C - 2\Delta \right) + \nu\left(\theta_A+\theta_B+\theta_C - \pi \right)$$
The partial derivative wrt $\theta_A$ is $$F_{\theta_A}=2\cos\theta_A+2\mu \cos 2\theta_A + \nu$$
Setting this and the other partial derivatives to zero gives a set of five equations to solve: $$\begin{align} 2\cos\theta_A+2\mu \cos 2\theta_A + \nu &=0 \\ 2\cos\theta_B+2\mu \cos 2\theta_B + \nu &=0 \\ 2\cos\theta_C+2\mu \cos 2\theta_C + \nu &=0 \\ \sin 2\theta_A+\sin 2\theta_B+\sin 2\theta_C &= 2\Delta \\ \theta_A+\theta_B+\theta_C &=\pi \end{align}$$
We'll prove that these equations imply two of the angles are the same.
For a contradiction, let's assume that they are in fact pairwise distinct. From the first two equations, we have $$\cos\theta_A+\mu \cos 2\theta_A = \cos\theta_B+\mu \cos 2\theta_B$$
so that $$\mu = \frac{\cos\theta_B- \cos \theta_A}{\cos2\theta_A-\cos 2\theta_B}=\frac{-1}{2(\cos \theta_A+\cos \theta_B)}$$
(note that, since $\theta_C \neq 0$, $\cos \theta_A + \cos \theta_B \neq 0$.) Likewise, $$\mu = \frac{\cos\theta_C - \cos \theta_B}{\cos2\theta_B-\cos 2\theta_C}=\frac{-1}{2(\cos \theta_B+\cos \theta_C)}$$
so that $$\cos \theta_A+\cos \theta_B = \cos \theta_B+\cos \theta_C$$
ie $\cos \theta_A = \cos \theta_C$
We assumed that $A\neq C$; so this means $\theta_A=-\theta_C$; but this contradicts the fact that the angles lie in $(0,\pi)$.
This contradiction proves that two of the angles must be the same - ie $ABC$ is isosceles.