Chain of Inscribed Circles

Construct two tangent circles Σ₁ and Σ₂ and the line L through their centers. We are going to inscribe into the crescent-shaped space between the circles a chain of pairwise tangent circles. The first one, α₀, has its diameter on L. The second, α₁, is tangent to α₀, and both Σ₁ and Σ₂, and so on. Let h_n and R_n denote the distance to L from the center and the radius of the circle α_n. Then

(The above proposition is commonly referred to as Pappus Circle Theorem or Pappus Chain Theorem. Pappus was of course unaware of the inversion. His proof is much more elaborate. He himself refers to it as "The Ancient Theorem" [Gaba].)

We can easily derive an additional result.

Assume the radius of the right semicircle is k times the radius of the left one. If the former is R_R, and the latter is R_L, we assume that

(2)	RR = k·RL.

Let R denote the radius of the big semicircle, then, since R = R_R + R_L, (2) implies that

(3)	RL = R/(1 + k) and RR = R·k/(1 + k).

For the sake of simplicity, let's choose the radius of the circle of inversion to be 2R. Then the left vertical line (this is the one that corresponds to the big semicircle) becomes tangent to both the big and the right semicircles as the common point of tangency lies then on the circle of inversion and is thus fixed. The right vertical line corresponds to the left semicricles. Introduce further notations. Let O_n be the center of a_n and P_n the projection of O_n on L. Let AB, CB, and AC denote the diameters of the semicircles. A is the center of inversion, AB = 2R, AC = 2R_L, CB = 2R_R, C' is the inversion of C, O' is the midpoint of BC'.

(2R)² = AB² = AC·AC'.

Thus

AC' = (2R)²/AC = 4R²/2RL = 2(1 + k)R.

And, since AO' = (AB + AC')/2,

AO' = (2 + k)R, BO' = kR.

So, for the same reason as (1),

hn/APn = 2nBO'/AO', or Rn/APn = BO'/AO'

And

Next, let's derive a generalization of Archimedes' formula:

The points of tangency of the common tangents through A of a pair of corresponding circles (like T_n and T'_n in the diagram) play a double role. On one hand, they are homologuous, corresponding under the implied homothety. On the other hand, they are antihomologous, mapping on each other under the inversion. More accurately

AT'n / ATn = R'n / Rn = Rk / Rn, and also AT'n · ATn = (2R)²

implying

(AT'n)² = 4R3k / Rn.

But (AT'_n)² = (2R + kR)² + (2n·kR)² - (kR)² which gives an equation for R_n, with the solution

Rn = Rk / (n²k² + k + 1).

which is (5). The combination of (4) and (5) gives

(6)	APn = R(k + 2) / (k²n² + k + 1).

Note that, t = 1/k, so that R_L = t·R_R then (5) becomes

(5')	Rn = Rt / (t² + t + n²).

References

1. M. G. Gaba, On a Generalization of the Arbelos, Am Math Monthly, Vol. 47, No. 1 (Jan., 1940), 19-24

Inversion

• Apollonian Circles Theorem
• Another Pair of Twins in an Arbelos
• Archimedes' Quadruplets
• Archimedes' Twin Circles and a Brother
• Bisectal Circle
• Chain of Inscribed Circles
• Circle Inscribed in a Circular Segment
• Feuerbach's Theorem: a Proof
• Four Touching Circles
• Hart's Inversor
• Inversion in the Incircle
• Circle Inversion: Reflection in a Circle
• Circle Inversion Tool
• Inversion with a Negative Power
• Peaucellier Linkage
• Polar Circle
• Poles and Polars
• Radical Axis of Circles Inscribed in a Circular Segment
• Steiner's porism
• Three Tangents, Three Secants

• Arbelos - the Shoemaker's Knife
• 7 = 2 + 5 Sangaku
• Another Pair of Twins in Arbelos
• Archimedes' Quadruplets
• Archimedes' Twin Circles and a Brother
• Book of Lemmas: Proposition 5
• Book of Lemmas: Proposition 6
• Chain of Inscribed Circles
• Concurrency in Arbelos
• Concyclic Points in Arbelos
• Ellipse in Arbelos
• Gothic Arc
• Pappus Sangaku
• Rectangle in Arbelos
• Squares in Arbelos
• The Area of Arbelos
• Twin Segments in Arbelos
• Two Arbelos, Two Chains

Copyright © 1996-2008 Alexander Bogomolny