Given mutually (externally) tangent circles $C_1,C_2,C_3$, let $C_n$ be the unique circle externally tangent to $C_{n-1}$, $C_{n-2}$, and $C_{n-3}$ for $n \geq 4$. Let $P_{\infty}$ be the point toward which the $C_n$'s tend, let $P_n$ be the center of $C_n$, and let $Q_n$ be the point of tangency between $P_{n}$ and $P_{n+1}$. What can be said about the asymptotic distribution of the unit vectors pointing from $P_{\infty}$ to $P_n$? What about the unit vectors pointing from $P_{\infty}$ to $Q_n$?

Although I would like the answer to be that these unit vectors are uniformly distributed over the unit circle, I see at least a couple of reasons to doubt this: first, I suspect that for certain special initial choices of $C_1,C_2,C_3$, the set of unit vectors that arise is finite, and second, the symmetry group governing Apollonian packings is the group of conformal maps, which does not preserve Lebesgue measure on the circle.

Still, I suspect that for generic choices of $C_1,C_2,C_3$, the unit vectors are asymptotically distributed according to a measure that is uniformly continuous with respect to Lebesgue measure on the circle, and I suspect that the measures that arise in this way admit a nice characterization, e.g., the set of all measures that are conformally equivalent to Lebesgue measure.

(This is a more refined version of my MathOverflow post Isotropy of Apollonian disk-packing . See also my follow-up question Three-dimensional Apollonian spirals .)

sauf erreuris $$ r_0 = \frac12\Bigl(1 + \sqrt{5} - \sqrt{2+\sqrt{20}}\Bigr) = 0.346014339\ldots $$ or equivalently the root $r_0^{-1}$ obtained by changing $-\sqrt\cdots$ to $+\sqrt\cdots$. Presumably this spiral yields irrational angles, and thus exact equidistribution. $\endgroup$ – Noam D. Elkies Nov 3 '14 at 23:44centers, but again the discrepancy disappears in the limit. $\endgroup$ – Noam D. Elkies Nov 5 '14 at 2:25