Could it get any more regular?

Let $O$ be the center of the circle and $\angle A = \theta$ . Then, $\angle BOC = 2\theta$ .

We note $\angle A + \angle B + \angle C = 1729\theta = 180^\circ$ , so $\theta = \frac{180^\circ}{1729}$ . Lining up adjacent triangles congruent to $\Delta BOC$ , we form our $n$ -gon. The angle sum at the center is then $2n\theta = 360^\circ$ .

Hence,

$n = \frac{360^\circ}{2\theta} = \frac{180^\circ}{180^\circ/1729} = \boxed{1729}$ .

$\angle A= \dfrac{\angle B}{864}~~\implies A+B+C= \dfrac{\angle B}{864} +B+B=180.\\\therefore B=\dfrac{180}{\dfrac{1}{864} + 2} ......(1)\\\therefore A=180-2*B....(2).\\\implies ~\text{BC substance an angle of 2A at the center......(3)}\\\text{ But this is one side of the n-gon }\therefore~using ~(1)~and~(2)~and~(3)\\ \large n=\dfrac{360}{2* \{180-2*\dfrac{180}{\dfrac{1}{864} + 2} \} }=\dfrac{1}{1-\dfrac{2}{\dfrac{1}{864} + 2} }=\boxed{ \color{#D61F06}{1729} }$

Hey @Jake Lai that's $\text{taxicab number}$ .