QA-Tf1: QA-Line Involution Center

 
 
General Information
In mathematics, an Involution, or an involutary function, is a function f that is its own inverse:
f(f(x)) = x for all x in the domain of f.
 
In EQF we will deal with the geometric notion of a “Line Involution”.
It occurs with pairs of points on a line, for whom the product of their distances from a fixed point P is a given constant. P is called the Center of Involution or Involution Center.
 
Let (A0,A1) and (B0,B1) be 2 pairs of points for whom the product of their distances from a fixed point O is a given constant c.
According to the definition of Line Involution: A0.O * A1.O = B0.O * B1.O = c.
When the points A0, A1 and the constant c are known we can calculate from a variable point B0 its mate B1. When we do the same calculation for B1 we find its mate B0. This makes the transformation an involutary function because f(f(P)) = P.
When BO=B1 we say that B0=B1 is a Double Point. There are 2 Double Points per Line Involution. Their Midpoint is the Center of Involution O.
  
Construction:  
QA-Tf1-Line-Involution-02
 
For a Line Involution let (A0, A1) and (B0, B1) be two pairs of points and ciA and ciB circles with these diameters.
The radical axis of ciA and ciB cuts the line in the Involution Center O.
A circle with center O perpendicular to ciA and ciB cuts the line in the two Double Points A2, B2.
Another construction can be found in Ref-19.
 
Desargues' Involution Theorem
A figure consisting of 4 points and their 6 connecting lines is called a (complete) Quadrangle. The 4 points are the vertices. The 6 connecting lines are the sides of the (complete) Quadrangle. Sides in a (complete) Quadrangle that have no vertices in common are called opposite sides.
Desargues’ Involution theorem states that the points of intersection of a line with the three pairs of opposite sides of a complete Quadrangle and a conic section circumscribed about the complete quadrangle form the pairs of an involution (see Ref-7).
Normally 2 pairs of points describe an involution. Desargues describes in his theorem 4 pairs of points all describing the same involution.
If we restrict ourselves to the 3 pairs of points generated from 3 sets of opposite sides we can deduce that when a line crosses a quadrangle it generates a unique Line Involution with an involution center and 2 double points (when real).
Consequently every line crossing a Quadrangle has an Involution Center and 2 Double Points (real or imaginary).
This line involution is a property of a Quadrangle (and not of a Quadrilateral as might be suspected).
In the QA-environment this will be called a QA-Line Involution.
 
Coordinates of the Involution Center and Double Points on some line L
Let L be a random line with barycentric coefficients (l:m:n).
Coordinates of Involution Center on L in CT-notation:
(m n p (m q + n r - l (q + r)) : l n q (l p + n r - m (p + r)) : l m (l p + m q - n (p + q)) r)
Coordinates of both Double Points on L in CT-notation:
(m n p (m q + n r) : -l m n p q - n W : -l m n p r + m W)
(m n p (m q + n r) : -l m n p q + n W : -l m n p r - m W)
where W = √(-l m n p q r (l p + m q + n r))
  
Properties:

 
Addendum
 
Calculations:
Let (A0,A1) and (B0,B1) be 2 pairs of points on a line that create a Line Involution with center O. Then according to the definition: d(A0,O) * d(A1,O) = d(B0,O) * d(B1,O).
QA-Tf1-Line-Involution-01
Let A2, B2 be the corresponding double points.
Now:
            d(A0,A2) = (a1*d1 - [d0*d1*a1*b1]) / (a1+b1)
            d(A0,B2) = (a1*d0 + [d0*d1*a1*b1]) / (a1+b1)
            d(B0,B2) = (b1*d0 - [d0*d1*a1*b1])  / (a1+b1)
            d(B0,A2) = (b1*d1 + [d0*d1*a1*b1]) / (a1+b1)
 
The center of involution O is given by:
            d(A1,O) = a1*d0 / (a1+b1)
            d(B1,O) = b1*d0 / (a1+b1)
and     d(A2,O) = d(B2,O) = [d0.d1.a1.b1] / (a1+b1)
 
Let (A0,A1) and (B0,B1) be 2 pairs of points on a line that create a Line Involution with center O.
Let A2 and B2 be the respective Double Points.
When (A0,A1) and B2 are known then we can calculate the position of A2:
Let      d3 = d(B2,A0) and d4 = d(B2,A1)
then   e0 = d(A2,A0) = (d3*d3-d3*d4)/(d3+d4),
and     e1 = d(A2,A1)  = (d4*d4-d3*d4)/(d3+d4).
 

 

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