Funky question. Given a regular n-gon P Let us let the inradius (apothem) length be r=1. Because then we will always have A = Area[circle] = π 1² = π (regardless of value of n). I'm not sure whether it makes it simplest to parameterize by sidelength a, inradius r, or longest diagonal, but these are related by: a = 2r tan(π/n) <=> r = a/2 cot(π/n) Denote by S the largest inscribed semicircle, and s as that semicircle radius. Then ratio A/B will always boil down to (π 1²)/(π/2 s²) = 2/s² So it comes down to determining s for each n. B = Area[semicircle] is going to be ad-hoc for small n. For small n, as the endpoints of the semicircle may not necessarily be on vertices, they might be on sides or indeed in the interior of the polygon. For large n (and as → ∞) as Michael points out, B ~ π/2, hence A/B → 2 Determining the largest semicircle boils down to picking the longest useable diagonal for a base (where to place the centre O). For large n the base will be the regular diagonal using floor(n/2) vertices. For small n the diagonal will not be on vertices (or maybe even sides), we have to consider (adhoc?) the maximum inradius as being the minimum distance from O to (possible tangency wrt) each of the sides. By symmetry, I think you will have to consider inradius wrt at most ceiling (n/2) sides. Without solving each ad-hoc case for n (I'll think about them), let's make the observation that polygon P is not as convex as its corresponding circumcircle is. The smaller n is, the worse (smaller) P is as an approximation to a circumcircle, hence the semicircle will have even more area chopped off (proportionately speaking) than the circle. Hence we would expect to have A/B ≥ 2 always, and decrease with increasing n, and tend to 2 as n→∞. [Exact answer to follow.] [n=3 case] The semicircle base (2s) must be placed along a side. If we take inradius as r=1, then sidelength a = 2 |tan(π/n)| = √3, and half-sidelength is √3/2. Semicircle base 2s < √3, and since the other two sides must be tangent to S, constructing the triangle with vertex angle π/3, opposite s and hypotenuse √3/2 ... s = √3/2 sin (π/3) = (√3/2)² = 3/4 Hence B = π/2 s² = 9π/32 Hence A/B = 2/s² = 32/9 ≈ 3.556 ; for n=3 [Not going to get back to this in time. To the others, thanks for adding detail. This bears a reasking - to look at the ratios in algebraic form, and see if there is any pattern to the sequence whatsoever.] @Todd and michael, look at the sequence for even n. Like I commented odd n implies some suboptimality. Surely the sequence for even n is strictly increasing? Probably the sequence for odd n is also strictly increasing?