A bullseye is a bullseye is a bullseye

You are playing darts with your friend, Apollonius, who has brought his own dartboard. However, this dartboard is somewhat … different. Instead of being a circle divided into concentric rings and sectors, the dartboard is a unit circle (i.e., with radius 1) that’s divided via an Apollonian gasket. In particular, this gasket is defined by two horizontally adjacent, congruent circles with radius 1/2.

But that’s not all! Every circle on the dartboard, no matter how small, is also its own Apollonian sub-gasket. Like the larger circle, every gasket, sub-gasket, sub-sub-gasket, etc., are defined by two horizontally adjacent, congruent circles with radii that are half the radius of their outer circle.

Now, when you throw a dart at this board, your score is the sum of the areas of every circle for which the dart lies inside or on the circumference. (Remember, the entire board is a unit circle.) What is the most a single dart can score?

Well as long as you hit the board at all you score $\pi$, so that's good. We note that if our dart hits the circumference of some circle, say with radius $r$, then there must be at lest some other circle, and in fact due to the nested Apollonian gaskets, infinitely many other circles of which this point is on the circumference. In particular, by recursion, if the dart lands at the point $z_0 + r_0 e^{i\theta}$ for some center, $z_0$, such that $\|z_0\|_2 \lt 1,$ and radius $r_0 \lt 1$, then there are infinitely many more circles $z_k + r_k e^{i\theta},$ where $r_k \leq \frac{1}{2^k}r_0,$ since all of the subcircles in the gasket have radius less than half of their outer circle, for some center $z_k$ satisfying $\|z_k - z_0\|_2 \lt r_0.$ Let's assume that the dart lands at the point of intersection between a circle of radius $r_1$ and a circle of radius $r_2,$ and that the smallest circle in the nested Apollonian gaskets that contains both of these circles is the unit circle, then the total score is $$S(r_1,r_2) \leq \pi + \sum_{n=0}^\infty \frac{\pi r_1^2}{4^n} + \sum_{m=0}^\infty \frac{\pi r_2^2}{4^m} = \pi \left( 1 + \frac{4}{3} \left( r_1^2 + r_2^2 \right) \right).$$

If we take any radii $r_1$, $r_2$ that appear in the Apollonian subgasket, then we get $$ \max S(r_1,r_2) \leq \max_{r_1, r_2} \pi \left( 1 + \frac{4}{3} \left(r_1^2 + r_2^2\right) \right) = \frac{5\pi}{3},$$ since $r_1, r_2 \leq \frac{1}{2}.$ On the other hand, if we get a bullseye, which sits exactly at the origin, then this sits in the circles indexed with centers at $z_k = (\textsf{sgn}(k) \frac{1}{2^{|k|}}, 0 )$ and radii $r_k = \frac{1}{2^{|k|}},$ for all $k \in \mathbb{Z}.$ So the score at the origin is $$S_0 = \sum_{k \in \mathbb{Z}} \pi \frac{1}{4^{|k|}} = \pi + 2\pi \sum_{k=1}^\infty \frac{1}{4^k} = \pi + 2\pi \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{5\pi}{3}.$$ Therefore, the bound is sharp and we must have that the maximum possible score of the dart, which occurs at a bullseye, is $\frac{5\pi}{3} \approx 5.235987756\dots.$