Soddy Circles and David Eppstein's Centers: What Are They?
A Mathematical Droodle

Explanation

Copyright © 1996-2008 Alexander Bogomolny

The applet suggests that for three circles externally tangent to each other, there exist two circles that touch all circles in the triplet. One of the circles touches the triplet internally, the other touches all of them externally. The circles bear the name of Frederick Soddy (1877-1956), a distinguished chemist and economist, a 1921 Nobel prize winner, who made the circles famous by publishing the result along with a poem "The kiss precise" in Nature, 137 (1936), p. 1021. The circles are correspondingly called the outer and the inner Soddy circles. Their centers are also known as the Soddy points of the triangle formed by the centers of the triplet.

The formula that links the radii of four circles that touch each other externally was known to Descartes [Pedoe, p. 158]. It is given in terms of the curvatures, bends in Soddy's terminology. For a circle, the curvature is the reciprocal of its radius, and Descartes' formula appears as ([Pedoe, p. 157, Coxeter, 1.57])

In 1826, the famous geometer Jacob Steiner rediscovered the formula. In 1842, it was proven independently by Philip Beecroft, an English amateur mathematician. Frederick Soddy's contribution appeared in 1936.

Thus, the Soddy circles have a long history, in the course of which they were under close attention of several exceptional people. This is then so much more remarkable that one interesting property of the configuration has been discovered as late as 2001. In 2001, D. Eppstein, of the Geometry Junkyard fame, published the following observation (see also a partial online version):

(1)	Four touching circles, when taken two by two, define two points of tangency and, therefore, a straight line. There are three such lines. The three lines are concurrent.

This is true for both inner and outer configuration, so there are two points of note, known now as Eppstein's points.

Eppstein derives his result as a particular case of a 3D configuration of four spheres. Any four mutually tangent spheres determine six points of tangency. The six points are naturally divided into three pairs of opposite tangencies, i.e. tangencies, in which one is defined by two spheres distinct from the pair of the spheres defining the other.

Lemma

[Altshiller-Court, p. 231].

The three lines through the opposite points of tangency of any four mutually tangent spheres in R3 are concurrent.

Proof

Let the four given spheres S_i, i = 1, 2, 3, 4, have centers O_i and radii R_i. Let ε_i = ±1/R_i, where the sign minus is taken if S_i contains the other three spheres, and "+" otherwise. Then the point of tangency t_ij between S_i and S_j can be expressed as

tij = (εiOi + εjOj)/(εi + εj).

This is a weighted average of the centers of the two spheres. A similar average for the centers of all four spheres

(2)	M = (ε1O1 + ε2O2 + ε3O3 + ε4O4)/(ε1 + ε2 + ε3 + ε4)

could also be represented as

M = ((ε1O1 + ε2O2)t12 + (ε3O3 + ε4O4)t34)/(ε1 + ε2 + ε3 + ε4),

which means that M lies on the line joining the opposite tangencies t₁₂ and t₃₄. By the same argument, it lies on the other two lines.

(1) follows from Lemma when the centers of the four spheres happen to be coplanar.

If one starts with a triangle ABC, then there exists a unique triplet C_A, C_B, C_C of pairwise tangent circles centered at the vertices of the triangle. Let a, b, c be the side lengths of ABC (a = BC, etc.) Then the radii of the circles C_A, C_B, C_C are given by

RA = s - a, RB = s - b, RC = s - c,

where s is the semiperimeter of ABC, s = (a + b + c)/2. The points of tangency t_AB, t_AC, and t_BC are exactly the points where the incircle of ABC touches its sides. The three lines At_BC, Bt_AC, and Ct_AB are concurrent at the Gergonne point Ge.

(3)	Ge = (εAA + εBB + εCC)/(εA + εB + εC)

because, for example,

(εAA + εBB + εCC)/(εA + εB + εC) = (εAA + (εB + εC)tBC)/(εA + εB + εC)

so that the point in (2) lies on the line joining A and t_BC, and similarly for the other two lines. (2) suggests that a relationship between M and Ge should not be unexpected. Indeed, among other things, Eppstein proves the following

Theorem

Let S be the center of the inner Soddy circle. Then S, M, and Ge are collinear.

Proof

The proof follows by combining (2) and (3):

(2')	M = (εAA + εBB + εCC + εSS)/(εA + εB + εC + εS)   = ((εA + εB + εC)Ge + εSS)/(εA + εB + εC + εS).

It's known that S, Ge, and I, the incenter of ABC are collinear. M therefore lies on the same line. By symmetry, the same is true of the outer Eppstein point.

Reference

1. N. Altshiller-Court, Modern Pure Solid Geometry, Chelsea, 2nd ed, 1964
2. H. S. M. Coxeter, Introduction to Geometry, John Wiley & Sons, 1961
3. H. S. M. Coxeter, The Problem of Apollonius, Am Math Monthly, 75 (1968), pp. 5-15
4. D. Eppstein, Tangent Spheres and Triangle Centers, Amer Math Monthly 108 (2001), pp. 63-66
5. A. Oldknow, The Euler-Gergonne-Soddy triangle of a triangle, Amer Math Monthly 103 (1996), pp. 319-329
6. D. Pedoe, Geometry: A Comprehensive Course, Dover, 1970

2D problems that benefit from a 3D outlook

1. 4 travellers
2. Desargues' Theorem
3. Soddy Circles and Eppstein's Points
4. Symmetries in a triangle
5. Three circles - #1
6. Three circles - #2
7. Three circles problem - #3

Desargues' Theorem

• Desargues' Theorem
• 2N-Wing Butterfly Problem
• Cevian Triangle
• Do You Speak Mathematics?
• Desargues in the Bride's Chair (with Pythagoras)
• Menelaus From Ceva
• Monge from Desargues
• Monge via Desargues
• Nobbs' Points, Gergonne Line
• Soddy Circles and David Eppstein's Centers
• Pascal Lines: Steiner and Kirkman Theorems II
• Pole and Polar with Respect to a Triangle
• The Lepidoptera of the Circles

Copyright © 1996-2008 Alexander Bogomolny