Area of minimum regular polygon given three vertices

Question

Related question: Regular polygon determined by three vertices

I have solved a problem that is related to the linked question. It boils down to the question "given three vertices of a regular polygon, what is the smallest possible area of a regular polygon these vertices are a part of?"

A possible solution is to find the circumcenter and the angles it forms with each of the vertices and obtain a lcm (as the linked question's accepted answer describes). Another option as described by this tutorial is to find the circumradius (Heron's formula + area in terms of circumradius) and the inner angles (law of cosines) of the triangle described by the three points, use them to find the number of sides as $\pi$ over the greatest common denominator of the three inner angles and use this information to calculate the area.

My question is why does taking the quotient of $\pi$ and the gcd of the inner angles of the triangle give the number of sides of the polygon? I've tested this approach and it does work but I cannot understand why.

Answer 1

Given three points ${\bf z}_i$ in the plane (say, coordinatewise) it is easy to determine the radius $r$ and center ${\bf c}$ of their circumcircle. It follows that the regular polygon of minimal area having these points as vertices (if there exists such a polygon at all) has area $<\pi r^2$. If the coordinates of the ${\bf z}_i$ are simple expressions containing $\sqrt{3}$, or so, you might be able to detect that they are in fact vertices of a dodecagon, but how should you be able to realize that your ${\bf z}_i$ are in fact vertices of a regular $45\,381$-gon? It follows that your problem, fascinating as it is, is in a way ill-posed.