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.