Can You Fling the Fractal Darts?

A dartboard is a unit circle. Inside it sit two congruent circles of radius $\tfrac12$, tangent to each other and to the rim; inside each of those, two circles of radius $\tfrac14$; and so on, the pattern repeating forever inside every circle. A dart’s score is the sum of the areas of every circle it lands inside of or on. What is the most a single dart can score?

The Fiddler, Zach Wissner-Gross, December 5, 2025(original post)

Solution

Aim for the exact centre of the board. The two radius-$\tfrac12$ circles are tangent there, so the dart is on both. Inside the right one, its two radius-$\tfrac14$ children are tangent at its centre $(\tfrac12,0)$, and the left of those children, centred at $(\tfrac14,0)$, passes through the board’s centre too. The same happens inside the left radius-$\tfrac12$ circle, so the centre lies on two radius-$\tfrac14$ circles. Repeating, the centre lies on two circles of radius $(\tfrac12)^k$ for every $k\ge1$, together with the unit circle. The score is $$\pi + 2\sum_{k=1}^{\infty}\pi\Bigl(\tfrac12\Bigr)^{2k} = \pi + 2\pi\cdot\tfrac13 = \boxed{\tfrac{5\pi}{3}}.$$

The computation

Build the circles themselves, on the $x$-axis: each circle of radius $r$ centred at $c$ spawns two children of radius $\tfrac r2$ at $c\pm\tfrac r2$. Sum the areas (in units of $\pi$, so $r^2$) of every circle that actually contains the board’s centre, pruning any branch that has moved off it. The total is $\tfrac53$.

from fractions import Fraction as F
import sys; sys.setrecursionlimit(10000)
def score(c, r, depth):                  # area/pi of centre-covering circles below (c, r)
    if depth > 200 or abs(c) > r: return F(0)         # off the centre -> prune
    return r*r + score(c - r/2, r/2, depth + 1) + score(c + r/2, r/2, depth + 1)
print(float(score(F(0), F(1), 0)))       # 1.6667 = 5/3  ->  score 5*pi/3

Extra Credit

Now the dart lands at a point chosen uniformly at random on the board. On average, what score does it earn?

Solution

A clean identity does the work. For a uniform point in the unit disk (area $\pi$), the expected score is $$E[\text{score}] = \frac1\pi\sum_{C}\operatorname{area}(C)^2 = \pi\sum_{C} r_C^{4},$$ since each circle $C$ lies inside the board, so integrating its indicator returns its own area. With $2^k$ circles of radius $2^{-k}$ at level $k$, $$\sum_{C} r_C^4 = \sum_{k=0}^{\infty} 2^{k}\bigl(2^{-k}\bigr)^4 = \sum_{k=0}^{\infty} 8^{-k} = \tfrac87,$$ so $$E[\text{score}] = \boxed{\tfrac{8\pi}{7}}\approx 3.59 .$$ (The source’s extra-credit value is behind its paywall; this is my own, for the self-similar two-circles-per-level board above.)

The computation

Throw the darts: build the full circle tree to a fine depth, scatter uniform points over the board, and for each sum the areas of every circle containing it. The average lands on $\tfrac{8\pi}{7}$.

import numpy as np
circles = []
def build(c, r, d):
    circles.append((c, r))
    if d < 12: build(c - r/2, r/2, d + 1); build(c + r/2, r/2, d + 1)
build(0.0, 1.0, 0)
rng = np.random.default_rng(0); N = 500_000
rad = np.sqrt(rng.uniform(0, 1, N)); ang = rng.uniform(0, 2*np.pi, N)
x, y = rad*np.cos(ang), rad*np.sin(ang); total = 0.0
for c, r in circles:
    total += np.pi*r*r*(((x - c)**2 + y*y) <= r*r).sum()
print(round(total/N, 4), round(8*np.pi/7, 4))   # ~3.59  3.5904