>>11608812Easy.
Construct an equilateral triangle of any size. Compute the perpendicular bisectors of the lines, and construct their intersection. Connect the intersection point with the endpoints of the equilateral triangle, and extend inefinitely. Take the radius of the circle you are working with, and construct a circle with that radius at the intersection point. The circle will intersect all three of the indefinitely long lines, forming three isosceles triangles. Because the indefinitely long lines were constructed from an equilateral triangle, the isosceles triangles must all have SAS congruency, hence, you constructed an equilateral triangle inside the circle you want to work with.
Bisecting the edges of this regular, inscribed polygon is simple enough, just connect the edge bisection points to the center point, and extend to the circumference. The new circumference points should be connected in sequence to the previous inscribed polygon's circumference points, constructing an inscribed, regular polygon of 2n sides (n is the number of sides of the previous regular, inscribed polygon).
Now that an arbitrary regular polygon of 3*2^n sides can be inscribed inside of the circle in question, one can simply convert this inscribed polygon to a circumscribed polygon by treating the endpoints of the circumscribed polygon as apothem points of the circumscribed polygon. In other words, you just take those lines emanating from the center, to the 3*2^n circumference points, and construct lines perpendicular to the radii, at the circumference points (this can be trivially done by letting the emanating lines continue for twice the radius of the circle, and then applying line bisection on those line segments). These tangent lines can be extended until the intersect neighboring tangent lines, and those intersections will be the points of the regular, circumscribed polygon of 3*2^n sides.
Computing the areas of the inner and outer polyhedra is trivial.
