Theorem: The sum of the curvatures of opposite circles in a Steiner ring of n circles with n even is constant. Proof: Consider the canonical n-ring C having concentric inner and outer circles, centred at the origin, with the first circle of the ring centred on the x-axis. Denote C_j the j-th circle of the ring, with j taken modulo n ; Ci , Co the inner and outer circles tangential to all C_j . Easily, C_j has centre ( cos(2pi j/n), sin(2pi j/n) ) , radius s = sin(pi/n) ; C_(j+n/2) has centre ( -cos(2pi j/n), -sin(2pi j/n) ) , radius s ; Ci has centre ( 0, 0 ) , radius 1-s ; Co has centre ( 0, 0 ) , radius -1-s (note negative sign). Any n-ring D equals C transformed by some Moebius inversion X , and conversely. We represent oriented circles and inversions in the plane by Lie-sphere (contact geometry) coordinate vectors with 5 homogeneous components: then X = [x, y, u, v, 0] ; C_j = [(1-t^2)/(1+t^2), 2*t/(1+t^2), s^2/2, 1-s^2/2, s] ; C_(j+n/2) = [-(1-t^2)/(1+t^2), -2*t/(1+t^2), s^2/2, 1-s^2/2, s] ; D_j = -X * C_j * X ; D_(j+n/2) = -X * C_(j+n/2) * X ; for some x,y,u,v , and appropriate t ; and `*' denotes product in the Clifford algebra Cl(3, 2) . The signed curvature of a circle D is given by (D[3] + D[4])/D[5] ; hence X increases the sum of opposite curvatures by factor ( (D_j[3] + D_j[4])/D_j[5] + (D_(j+n/2)[3] + D_(j+n/2)[4])/D_(j+n/2)[5] ) / ( (C_j[3] + C_j[4])/C_j[5] + (C_(j+n/2)[3] + C_(j+n/2)[4])/C_(j+n/2)[5] ) which reduces to -(s^2 u^2 + 2 s^2 u v + s^2 v^2 - u^2 - 2 u v - v^2 - x^2 - y^2) / (u^2 - v^2 + x^2 + y^2) . This is independent of t , and hence of j . The sum is plainly constant for C , increases by a constant factor for D , therefore remains constant wrt j . QED A similar computation for inner and outer curvatures shows that the same increase applies to their sum, which equals the first sum times 1/s^2 - 1 . The result generalises immediately to m-fold rings with n replaced by n/m . In fact, n plays no essential part in the main computation, which relies only C ring circles having centres lying on the unit circle. Their parametrisation by t is a convenient ruse avoiding complications in the CAS program. Fred Lunnon On 8/9/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
On 8/8/14, Bill Gosper <billgosper@gmail.com> wrote:
If the four curvatures are k1, k2, k3, k4, and the bounding circles are k0 and k5, then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of signs to k0 and k5. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun