Apollonian gasket

Example

Start with four circles, one enclosing the other three, that are pairwise tangent at six distinct points. The three inner circles and the outer circle bound four curvilinear triangles. Into each such curvilinear triangle, insert the unique circle inside it tangent to its three sides. This creates three new curvilinear triangles. Repeat forever. The Apollonian gasket is the family of all circles obtained, or the closure of their union. Apart from the outer circle, which encloses the others, the circles have disjoint interiors, so they form a circle packing. For the curvatures, take the outer circle with negative sign. By Descartes' circle theorem, four pairwise tangent circles with curvatures $k_1, k_2, k_3, k_4$ satisfy $(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2)$. Check with the start $(-1, 2, 2, 3)$: an outer circle of radius $1$, two circles of radius $\tfrac12$ and one of radius $\tfrac13$; then $6^2 = 36 = 2(1 + 4 + 4 + 9)$. Each new circle replaces one curvature $k_4$ by $2(k_1 + k_2 + k_3) - k_4$, so in this gasket all curvatures are integers.