I hope Henry or someone is collecting our various complex number geometricks e.g., PolygonArea[L_List] := Total[MapThread[Im[#1\[Conjugate]*#2] &, {L, RotateLeft[L]}]]/2 line segment intersect Interseg[z1_, z2_, z3_, z4_] := Block[{z12 = z1 - z2, z34 = z3 - z4}, (z34*Im[z2*z1\[Conjugate]] + z12*Im[z3*z4\[Conjugate]])/Im[z34*z12\[Conjugate]]] incenter, circumcenter, inradius, ... . For the Clifford circle theorems I needed second intersection: Given centers of circles though the origin z1, z2, secondintersection[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2] Is this obvious to anybody? I had to derive it with 8th grade analytic geometry. --rwg
Can I just interject that such computations involving circles (and spheres in higher dimension) are all special cases of the "Generalised Apollonian Problem", which has a complete formal solution in terms of the Clifford algebra based on pentacyclic coordinates for plane contact (Lie-sphere) geometry. Looking ahead, can anyone come up with an algebraic proof of Euler-Chapple: (R - 2r) R = d^2 where R, r, d denote circumradius, inradius (or exradius), displacement of centres resp. of a plane Euclidean triangle? My sledgehammer attempts have so far met with humiliating defeat ... WFL On 1/28/16, Bill Gosper <billgosper@gmail.com> wrote:
I hope Henry or someone is collecting our various complex number geometricks e.g., PolygonArea[L_List] := Total[MapThread[Im[#1\[Conjugate]*#2] &, {L, RotateLeft[L]}]]/2
line segment intersect Interseg[z1_, z2_, z3_, z4_] := Block[{z12 = z1 - z2, z34 = z3 - z4}, (z34*Im[z2*z1\[Conjugate]] + z12*Im[z3*z4\[Conjugate]])/Im[z34*z12\[Conjugate]]]
incenter, circumcenter, inradius, ... .
For the Clifford circle theorems I needed second intersection: Given centers of circles though the origin z1, z2,
secondintersection[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2]
Is this obvious to anybody? I had to derive it with 8th grade analytic geometry. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I feel that using fractional ought to help somehow. Consider a circle C_a through 0 with center at the point a in C. What does it become upon application of 1/z ? We know fractional linear transformations take lines and circles to lines and circles. 1/C_a "passes through oo", so must be a straight line. Which one? The point 2a is the farthest point from 0 of C_a, so it follows that 1/(2a) is the farthest point from oo of 1/C_a, i.e. the nearest to 0. Thus 1/C_a is the line L_a passing through 1/(2a) perpendicular to 1/(2a), i.e., L_a(t) = 1/(2a) + t a*, t in R , where z* denotes conj(z). Now let a and b denote z_1 and z_2, respectively. (*) We exclude the cases Ka + Lb = 0 for some K, L in R as exceptional. Applying 1/z to both circles C_a and C_b, we get the lines L_a(s) = 1/(2a) + s a*, s in R and L_b(t) = 1/(2b) + t b*, t in R whose intersection is at, say, (s,t) = (S,T). At this stage I don't know a painless way to get the complex intersection point of L_a and L_b. But when you find it, say at the point J(a,b), the original second intersection will be at its reciprocal, 1/J(a,b). Condition (*) above suggests there should be a denominator of the form a*b - ab* — or someting like that! — in the expression, since that being 0 is equivalent (maybe after adjustment) to (*). —Dan
On Jan 28, 2016, at 2:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
. . . For the Clifford circle theorems I needed second intersection: Given centers of circles though the origin z1, z2,
secondintersection[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2]
Is this obvious to anybody? I had to derive it with 8th grade analytic geometry.
On Thu, Jan 28, 2016 at 2:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
I hope Henry or someone is collecting our various complex number geometricks e.g., PolygonArea[L_List] := Total[MapThread[Im[#1\[Conjugate]*#2] &, {L, RotateLeft[L]}]]/2
line segment intersect Interseg[z1_, z2_, z3_, z4_] := Block[{z12 = z1 - z2, z34 = z3 - z4}, (z34*Im[z2*z1\[Conjugate]] + z12*Im[z3*z4\[Conjugate]])/Im[z34*z12\[Conjugate]]]
incenter, circumcenter, inradius, ... .
For the Clifford circle theorems I needed second intersection: Given centers of circles though the origin z1, z2,
secondintersection[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2]
Is this obvious to anybody? I had to derive it with 8th grade analytic geometry. --rwg
It was obvious, but only to (young eavesdropper) Zack Chroman: In[568]:= Abs[z1 - x] == Abs[z1]; In[569]:= Abs[z2 - x] == Abs[z2]; In[572]:= Abs[w_] :> Sqrt[w*Conjugate@w] In[573]:= {%568, %569} /. % At this point, Mathematically spontaneously and bizarrely switched contexts! Out[573]= {Sqrt[(-Global`x + Global`z1) (-Conjugate[Global`x] + Conjugate[Global`z1])] == Sqrt[Global`z1 Conjugate[Global`z1]], Sqrt[(-Global`x + Global`z2) (-Conjugate[Global`x] + Conjugate[Global`z2])] == Sqrt[Global`z2 Conjugate[Global`z2]]} In[574]:= $ContextPath Out[574]= {"Parallel`Protected`", "Parallel`Developer`", "Parallel`", "System`"} In[576]:= $Context = "Global`" Out[576]= "Global`" In[577]:= %573 Out[577]= {Sqrt[(-x + z1) (-Conjugate[x] + Conjugate[z1])] == Sqrt[ z1 Conjugate[z1]], Sqrt[(-x + z2) (-Conjugate[x] + Conjugate[z2])] == Sqrt[z2 Conjugate[z2]]} Squaring everything In[601]:= %577 /. a_ == b_ :> a^2 == b^2 Out[601]= {(-x + z1) (-Conjugate[x] + Conjugate[z1]) == z1 Conjugate[z1], (-x + z2) (-Conjugate[x] + Conjugate[z2]) == z2 Conjugate[z2]} In[293]:= Solve[{(-x + z1) (-Conjugate[x] + Conjugate[z1]) == z1 Conjugate[z1], (-x + z2) (-Conjugate[x] + Conjugate[z2]) == z2 Conjugate[z2]}, {x, Conjugate@x}] Out[293]= {{x -> 0, Conjugate[x] -> 0}, {x -> (z2 Conjugate[z1] - z1 Conjugate[z2])/( Conjugate[z1] - Conjugate[z2]), Conjugate[x] -> (-z2 Conjugate[z1] + z1 Conjugate[z2])/(z1 - z2)}} QED --rwg Failure to square leaves Solve with only x->0, and Reduce still cogitating after most of an hour.
participants (3)
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Bill Gosper -
Dan Asimov -
Fred Lunnon