Complex Descartes Circle Theorem

We present a short proof of Descartes Circle Theorem on the “curvaturecenters” of four mutually tangent circles. Key to the proof is associating an octahedral configuration of spheres to four mutually tangent circles. We also prove an analogue for spheres. It can be traced back to at least Descartes that four mutually tangent circles have curvatures (reciprocals of radii) satisfying the relation (a+ b+ c+ d) = 2(a + b + c + d). (1) The Monthly has published several papers concerning this fascinating topic: [1, 3, 4, 5, 6]. It was only in 2001 [3] that is was noticed, and proved, that the “curvature-centers” (curvature times center where the center is considered a complex number) satisfy the same relation. We present a short proof of this result (Theorem 1) and an analogous version for spheres (Corollary 1). For the purposes of this paper, a sphere will alway be contained in the halfspace C × [0,∞) and be tangent to the complex plane. Let S(z, r) denote the sphere with radius r tangent to C at z. It is obvious that S(z, r) and S(w, s) are tangent to each other if and only if |z − w| = 4rs. It is also immediate that given any three points z1, z2, z3 ∈ C, there are unique numbers r1, r2, r3 such that the spheres S(zi, ri) are mutually tangent. In particular, if {i, j, k} = {1, 2, 3} then ri = |zi − zj| · |zi − zk| 2|zj − zk| . (2) We say that two circles are orthogonal if they intersect at right angles; see Figure 1. Lemma 1. Let C1, C2 be two orthogonal circles which intersect at w1, w2 and that have curvatures c1, c2 and centers z1, z2 respectively. Let k1, k2 be the curvatures of any two tangent spheres tangent to C at w1, w2 respectively. Then (a) k1k2 = 4 |w1 − w2|2 = c 1 + c 2