> It seems easier to construct rational point polygons than rational side length > > polygons, for the regular and convex it is all inside the triangle. (It is > > easy to construct the rational divisions of the angles.) Then, for the chord > > and line length, it is always the chords of the radius for the side length to > > match 1-1 with arc length, for just a rat
ional side length edge. Then for > > the rest of the edges it is whether the rest of the circle, worked out from the > > arc length, with have these rational edges. The idea is to treat it from 0 to > > 2pi, around the circle, but also -pi to pi. 0 to 2pi is counter-clockwise, ccw > > (right hand rule), -pi to pi is clockwise, or ccw. This is where the >
> properties of all the vertices of the polygon are that they are on the circle > > for convex vertices (for convex polygons) and if internal then concave. It > > seems obvious there are rational side length pairs, connecting any two points, > > eg of the square of the line through them as a diagonal. So, concave polygons > > with rational side lengths, can be constructed from rational convex polygons, > > here as to whether its possible to construct, for an irrational length, two > > rational lengths connected by a third point here inside the circle, of a > > rational length through the third point between the vertices otherwise at an > > irrational distance. Then the idea is that there are those, concave polygons > > bounded by a circle with only rational side lengths for any collection of > > vertices on the circle, then that from actual rational side lengths, they have > > all of regular geometry around them then, the regular n-gons. Then the idea > > is to use constructions of arc length, and the convex, but that the concave is > > the case for all rational lengths in the geometry, then as to that the > > cyclotomic fields are regular with usual reversibility in cyclotomic fields, > > here for 0 to 2pi or -pi to pi. Given two points, is it possible to construct > > a third point of rational distance to each, compare ed to a third distance fixed > > as one? It seems so where it is less than two. Given two points that are > > irrational, is it possible to construct a point at rational distances from each > > endpoint, here without distance fixed? Here it is so if above, for whatever > > rational distance there would be, that the path would be of rational length. > > Then though as to the rational division of the angles, that is of the > > isosceles, and the base being regular in the isosceles where it is irrational, > > has whether it is then of the irrational components in the angle formulae, to > > divide those out, then for building the convex polygons on the circle with > > those. Then the side length, of the convex, to each be an integer in relation > > to the others as some integers, as scaled from 1.0, for them to be rational, it > > divides the arc length with the same ratios. The arc length is 2pi radians. > > What it is there is that the polygon on the circle is the outside of the > > connected set of all the vertices with the vertices on the circle, all the > > edges are inner. Here for the outermost edges as there are, these to be > > regular or as to rational would be then that as connected, the points have > > usual constructions to all the other points, here up in the cyclomatic fields >
> of those, the vertices of the polygon on the circle, as connected set of edges, > > with that there are outer edges so the polygon is convex, then that it is on > > the circle so the connections conserve a maximum or inequality, here of the > > angles and thus distances. Then like Gauss is finding in the Mersenne, the > > prime and co-prime in the cyclomatic may have tractable forms for rational > > (PRUC one point) n-gons. > > > > For general interest in geometry, there is the regular inscribed n-gon. Is the
> > line segment the two-sided case? It is always rational. There are a lot of > > ways to find many of these. It's nice to type these, I wrote that an hour ago > > then just now add a few words. Mostly though it's all one spew. And me typing > > it is the spell-check. The polygon's vertices that are each irrational, adding > > to those more vertices of rational distance, the result can be a polygon of > > only rational lengths. Then obviously the original vertices at at most one not > > originally rational, that's though a definition because the original n > > irrational points just have one that can surely be made rational as zero by > > subtracting it, but also 4-5 them here for the 2n-1, and triangle points, then > > also for 4 and 5 and line points.