# Spirals in Apollonian circle-packings

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 .)

• All such configurations are inversively equivalent, and any inversion is angle-preserving and a local homothecy. So it's enough to consider the case that $C_1,C_2,C_3$ have radii chosen so that the radii of $C_1,C_2,C_3,C_4$ are in geometric progression, because then by induction the same is true for all the $C_n$, whence $P_n$ and $Q_n$ lie on actual spirals. If I did this right the common ratio of this progression is [cont'd] Nov 3 '14 at 23:43
• [cont'd] a positive root of $r^2 + r^{-2} = 2(r+r^{-1}+1)$, which sauf erreur is $$r_0 = \frac12\Bigl(1 + \sqrt{5} - \sqrt{2+\sqrt{20}}\Bigr) = 0.346014339\ldots$$ 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. Nov 3 '14 at 23:44
• I should point out a subtlety that some readers may have missed: conformal maps are only locally angle-preserving, so the angular distribution of the unit vectors in the general case cannot be obtained by applying a rotation or other simple map to the unit vectors in the special case that Noam proposes. However, since the points $P_n$ and $Q_n$ all approach $P_{\infty}$, and since the inversive map preserves angles in the vicinity of $P_{\infty}$, we can ignore this distortion for purposes of the asymptotic angular distribution of the vectors. Nov 5 '14 at 2:11
• I like Noam's argument. Can anyone complete the proof by showing that the angle in question is irrational? Nov 5 '14 at 2:13
• Thanks. Meanwhile I did verify this, and will post some more details soon. For now you can zoom in on the picture at math.harvard.edu/~elkies/mo186137.pdf to see a few iterations of the limiting configuration. BTW there's the additional small subtlety that conformal maps don't preserve circle centers, but again the discrepancy disappears in the limit. Nov 5 '14 at 2:25

$$k=\varphi + \sqrt{\varphi} \approx 2.89005\ ,$$ where φ is the golden ratio. k and its reciprocal satisfy the equation $$(1+x+x^2+x^3)^2=2(1+x^2+x^4+x^6)\ .$$ The centres of the circles in the sequence lie on a logarithmic spiral. Viewed from the centre of the spiral, the angle between the centres of successive circles is $$\cos^{-1} \left( \frac {-1} {\varphi} \right)\ .$$