Shouldn't QNED ("Quod Non Erat Demonstrandum") be reserved for reminding readers that the previous assertion was NOT what we were trying to prove? :-) Jim On Tuesday, August 19, 2014, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Readers with a limited attention span are invited to restrict attention to numbered formulae below.
At some stage sketched proofs, less obvious sample chains, pictures are envisaged ...
WFL
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Steiner Chain Constraints _________________________
"Circles" (more accurately regarded as open discs) shall be oriented, and their curvatures equipped with signs: negative signs indicate circles inside-out (complemented discs). Lines and points, as limiting cases of circles with curvature 0 and oo , will be ignored for the present. A "Steiner chain" comprises a set of circles touching given frontier circles A,B , and indexed by j such that member j touches also j-1 and j+1 .
Circles touching in a chain have signs matching just when external to one another, their corresponding discs disjoint: this ensures in particular that the union of chains associated with given frontier constitutes a unique continuous regulus. [Technically such circles are "anti-tangent", their touching not preserved under general transformations of contact geometry.] The distinction may seem frivolous: however, failing to respect signs leads rapidly to horrible confusion.
In general j is defined only modulo n , and n and j may be real (elliptic, eg. annular), zero (parabolic, eg. arbelos), or imaginary (hyperbolic with intersecting frontiers). A classical "closed" chain has both n and j natural; or relaxation to n = p/q rational corresponds to a chain of p circles winding q times around some point.
An elliptic chain is the (Moebius) inversion of a canonical chain with frontier circles concentric at the origin, and chain members centred on the unit circle: j can be defined by the pre-image of member j having its centre at argument 2pi j/n , for some value of parameter n . [Steiner's porism asserts that n is independent of continuous translations of j , a consequence of rotating the canonical chain with the same value n .]
Below, frontier curvatures are denoted kA,kB ; while given some arbitrary value of j , curvatures of a consecutive chain of anti-tangent members j+1, j+2, j+3, ..., are abbreviated to k1, k2, k3, ... .
** Theorem ** Suppose n fixed, and denote t == tan(Pi/n) . Two frontier and any three consecutive member curvatures satisfy constraints kA + kB = b k1 + 2(b-1) k2 + b k3 , (1) where b = (t^2 + 1)/2 ; and (kA - kB)^2 = u k2^2 + v ( k1 k2 + k3 k1 + k2 k3 ) , (2) where u = (t^2 - 3)(t^2 + 1), v = (t^2 + 1)^2 .
Proof: Routine using contact (Lie-sphere) transformations, plane pentacyclic coordinates and geometric algebra for inversion --- some other time! QNED
These constraints and their integer j-shifts determine all curvatures in terms of n and k1,k2,k3 . Indeed for n = p/q <> 0 they evidently constitute a basis for the polynomial ideal of possible chain constraints: for p >= 2 integer, (2) alone together with p-2 shifts of (1) suffice. Treating (1) as a linear recurrence indexed by j , similar relations may be obtained involving (non-consecutive) subsets of indices.
** Examples ** Basis n = 3 find t^2 = 3 , b = 2 , u, v = 0, 16 ; { - (kA + kB) + 2 (k1 + k2 + k3) , - (kA - kB)^2 + 16 ( k1 k2 + k3 k1 + k2 k3 ) } . Essentially this is Soddy's identity, interpreted as an equation with roots kA,kB . Sample (annular closed) chains: [kA, kB, k1, k2, k3] = [ -1, 3; 2, 3 ] , passim.
Basis n = 4 find t^2 = 1 , b = 1 , u, v = -4, 4 ; { - (kA + kB) + k1 + k3 , - (kA + kB) + k2 + k4 , - (kA - kB)^2 - 4 k2^2 + 4 ( k1 k2 + k3 k1 + k2 k3 ) } . Besides the dihedral group D_2n acting on j , and transposing A,B , here any of 3 pairs of disjoint circles may be chosen as bounding, yielding 128 symmetries in all; the constraints thus generated still lie within the theorem ideal. Sample chains: [ -1, 7; 2, 2, 4, 4 ] , [ -2, 16; 3, 8, 11, 6 ] .
Basis n = 6 find t^2 = 1/3 , b = 2/3 , u, v = -32/9, 16/9 ; { - (kA + kB) + (2/3) (k1 - k2 + k3) , - (kA + kB) + (2/3) (k2 - k3 + k4) , - (kA + kB) + (2/3) (k3 - k4 + k5) , - (kA + kB) + (2/3) (k4 - k5 + k6) , - (kA - kB)^2 - (32/9) k2^2 + (16/9) ( k1 k2 + k3 k1 + k2 k3 ) } . You get the general idea ... Sample chain: [ -2, 10; 9, 3, 6, 15, 21, 18 ] .
Basis n = 5 : find tau == (1+sqrt5)/2 , t^2 = 7 - 4 tau , b = 2/tau^2 , u, v = -tau, 1 ; { - (kA + kB) + (2/tau^2) (k1 + k3) - (2/tau^3) k2 , - (kA + kB) + (2/tau^2) (k2 + k4) - (2/tau^3) k3 , - (kA + kB) + (2/tau^2) (k3 + k5) - (2/tau^3) k4 , - (kA - kB)^2 - tau k2^2 + ( k1 k2 + k3 k1 + k2 k3 ) } . Which might come as a surprise if you casually assumed rational coefficients; no use expecting samples with rational curvatures!
** Corollary ** This elegant " n-free" constraint (k1 - k4)(k4 - k3) = (k2 - k5)(k3 - k2) (3) arises via eliminating b or u/v from consecutive copies of (1) or (2).
Now suppose we are given only k1,k2,k3,k4 , with n unspecified. In general t is recovered via d == (k1 - k4)/(k2 - k3) = (3 - t^2)/(1 + t^2) , whence (involutorily!) t^2 = (3 - d)/(1 + d) = 4(k2 - k3)/(k2 - k3 + k1 - k4) - 1 ; and finally n = Pi/arctan(t) , real or imaginary.
Expressing b,u,v of the theorem in terms of t^2 , then eliminating t^2 as above, yields these n-free constraints involving kA,kB . ** Corollary ** (kA + kB) (k1 + k2 - k3 - k4) = 2 ( k2 (k2 + k4) - k3 (k3 + k1 ) , (4) (kA - kB)^2 (k1 + k2 - k3 - k4)^2 = 16 (k2 - k3) ( k2^2 (k3 + k4) - k3^2 (k1 + k2) ) . (5) [For a closed chain their integer shifts are plainly still complete; however they no longer constitute a basis, since no linear constraint (1) can lie in this ideal.]
Geometric features of a chain, via t^2 > 0 for real n --- -1 < d < 0 or 0 < d < 3 elliptic: A,B disjoint, incl. annular & closed; d = 3 parabolic: Pappus arbelos, A,B anti-tangent; 3 < d < oo or oo < d < -1 hyperbolic: A,B intersecting; d = oo or d = 0 impossible: eg. [1,1,1,2] ; (6) d = 0/0 incomplete: extremum at k_(5/2) , incl. A,B concentric.
Fred Lunnon [19/08/14]
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