HYACINTHOS 2704

• Dear Richard and friends,
  
  I do not know if I interpret correctly; but it seems to me that a
  reasonable clock pivot P point is one such that
  
  angle BPC : angle CPA : angle APB = a : b : c.
  
  Note that with s = (a+b+c)/2,
  
  angle BPC = (a/s)pi etc.
  
  This clock pivot point then has homogeneous barycentric coordinates
  
  PB.PC.sin BPC : PC.PA.sin CPA : PA.PB.sin APB
  = (1/PA).sin(a/s)pi : (1/PB).sin (b/s)pi : (1/PC)sin(c/s)pi.
  
  This can be regarded as the ``product'' of the (or an*) isodynamic
  point and the point
  
  (sin (a/s)pi : sin (b/s)pi : sin (c/s)pi).
  
  * For practical purposes one uses the isodynamic point that is
  interior to triangel ABC. By using the other isodynamic point, we get
  the inverse of this (interior) clock pivot point with respect to the
  circumcircle.
  
  For a right triangle with sides a,b,c, the isodynamic point is
  
  (a^2(b^2+c^2-a^2).sqrt 3 + 4 triangle) : ... : ...)
  ~(a^2b^2.sqrt 3) + a^2.ab : a^2b^2.sqrt 3 + b^2.ab : c^2.ab)
  ~(ab.sqrt 3 + a^2 : ab.sqrt 3+b^2 : c^2)
  =(a(a+b.sqrt 3) : b(b+a.sqrt 3) : c^2).
  
  If a:b:c = 3:4:5, this is
  
  (3(3+4.sqrt 3) : 4(4+3.sqrt 3) : 25),
  
  and the other point has very simple coordinates
  
  (sin pi/2 : sin (2 pi)/3 : sin (5 pi)/6)
  ~(2 : sqrt 3 : 1).
  
  The clock pivot point for the 3:4:5 triangle is therefore
  
  (6(3+4.sqrt 3) : 4.sqrt 3(4+3.sqrt 3) : 25)
  ~(6(3+4.sqrt 3) : 4(9+4.sqrt 3) : 25)
  
  If we normalize these so that the sum of the coordinates is 1, we get
  
  (6(-243+196.sqrt 3),44(21-4.sqrt 3),25(79-40.sqrt 3))/(1441).
  
  If we choose (4,0) for A, (0,3) for B, and (0,0) for C [so that 9
  o'clock is at C, 12 o'clock at B, and 5 o'clock at A], this clock
  pivot point is
  
  (24(-243+196.sqrt 3),132(21-4.sqrt 3))/(1441)
  ~(1.60692,1.28902).
  
  Best regards
  Sincerely
  Paul Υiu