Math 2.0

Geometry Triangles and circles Example

Steiner chain

Builds on Circle

Example

Let α\alpha and β\beta be two circles in the plane with β\beta inside α\alpha and no common point. A Steiner chain for α\alpha and β\beta is a cyclic sequence of n≥3n \ge 3 circles γ1,…,γn\gamma_1, \ldots, \gamma_n with two properties. Each γi\gamma_i is tangent to α\alpha from inside and to β\beta from outside. Each γi\gamma_i is tangent to γi+1\gamma_{i+1}, where γn+1=γ1\gamma_{n+1} = \gamma_1.

Take α\alpha of radius 33 and β\beta of radius 11, both centred at the origin. Let γk\gamma_k be the circle of radius 11 centred at 2(cos⁡(kπ/3),sin⁡(kπ/3))2(\cos(k\pi/3), \sin(k\pi/3)) for k=1,…,6k = 1, \ldots, 6. The centre of γk\gamma_k is at distance 2=3−12 = 3 - 1 from the origin, so γk\gamma_k touches α\alpha from inside. It is also at distance 2=1+12 = 1 + 1, so γk\gamma_k touches β\beta from outside. Two consecutive centres are at distance 2⋅2sin⁡(π/6)=22 \cdot 2 \sin(\pi/6) = 2, the sum of the radii, so γk\gamma_k and γk+1\gamma_{k+1} touch. So these six circles form a Steiner chain.

Remarks

For concentric circles of radii R>rR > r, a chain of nn circles that goes once around β\beta exists exactly when sin⁡(π/n)=(R−r)/(R+r)\sin(\pi/n) = (R - r)/(R + r). Steiner's porism says that if one chain closes up for α\alpha and β\beta, then a chain closes up starting from any first circle tangent to both. The proof uses an inversion in a circle that sends α\alpha and β\beta to concentric circles, where rotation moves one chain to any other.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Steiner_chain
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