Mazels on your wedding

The larger regular hexagon in the diagram below has a side length of $1.$ What is the side length of the smaller regular hexagon?

So let's first note that if the side length of the larger hexagon is $1,$ then the diameter of the circumscribing circle must be $2,$ thus it has a radius of $R = 1.$ This can be seen by noting that a regular hexagon can be divided into six equilateral triangles.

Assuming that both the larger hexagon and the circumscribing circle are centered at the origin, the upper semicircle is then given by point of the form $y = \sqrt{1-x^2}.$ In particular, let's assume that the side length of the smaller hexagon is $s,$ then the points where the smaller hexagon and the circumscribing circle intersect are $(\pm \frac{s}{2}, \sqrt{1 - \frac{s^2}{4}}).$

Again, noting that the regular hexagon can be broken into size equilateral triangles, we see that the height of the smaller hexagon is $s\sqrt{3},$ while half of the height of the larger hexagon is $\frac{\sqrt{3}}{2}.$ Therefore, the upper boundary of the smaller regular hexagon is on the line $y = \frac{\sqrt{3}}{2} + s \sqrt{3}.$ So, in particular, we have $$\sqrt{1 - \frac{s^2}{4}} = \frac{\sqrt{3}}{2} + s \sqrt{3}.$$

Squaring both sides we get $$1 -\frac{s^2}{4} = 3 \left( s + \frac{1}{2} \right)^2 = 3s^2 + 3s + \frac{3}{4},$$ or equivalently $$0 = \frac{13}{4} s^2 + 3s - \frac{1}{4} = \frac{13}{4} \left( s^2 + \frac{12}{13} s - \frac{1}{13} \right) = \frac{13}{4} (s + 1) \left(s - \frac{1}{13}\right),$$ so choosing the positive solution, we must have that the smaller hexagon has side length $\hat{s} = \frac{1}{13}.$

Generically, we can solve for any arbitrary number of additional vanishingly small hexagons by setting $s_1 = \hat{s} = 1/13$ as above, and setting up the recurrence relationship $$\sqrt{1 - \frac{s_i^2}{4}} = \sqrt{3} \left( \frac{1}{2} + \sum_{k=1}^i s_k \right),$$ where the side length of the $i$th hexagon is denoted by $s_i.$ We can square and solve this relationship to find that the side lengths of the next two smaller hexagons are $s_2 = \frac{-90 + \sqrt{8113}}{169} = 4.27179\cdot 10^{-4}$ and $s_3 = 1.31695\cdot10^{-18},$ respectively.