QLTf1: ClawsonSchmidt Conjugate
The ClawsonSchmidt Conjugate of a point P is a transformation of a point P in such a way that each defining line of the Reference Quadrilateral is transformed in the circumcircle of the triangle formed by the other 3 lines of the Reference Quadrilateral.
It is a conjugate because applying two times this transformation ends up in the original point.
Origin of the conjugate
John Wentworth Clawson suggested an inversion transformation wrt some circle with the Miquel point as center (see Ref22 and Ref40). It was Eckart Schmidt who gave a very useful form to this inversion by modifying it in this way:
In each quadrigon of a quadrilateral we have 4 points A, B, C, D in this order. Let M be the Miquel Point. Now the Angle Bisectors of <AMC and of <BMD are the same line. See Ref15d. Moreover this line is the same (invariant) for each QLQuadrigon. Let this line be called the 1st Steiner Axis. Let the 2nd Steiner Axis be the line perpendicular at the 1st Steiner Line in M.
Let the circle with Circumcenter M and with radius = √[MA.MC] = √[MB.MD] be called the Schmidt Circle.
The ClawsonSchmidt Conjugate of a point P is the Inversion wrt the Schmidt Circle of the Reflection in the 1st Steiner Line.
Moreover the reflection and inversion can be performed in reversed order.
The result is a conjugation that transforms each defining line of the Reference Quadrilateral into the circumcircle of the triangle formed by the other 3 lines of the Reference Quadrilateral.
Relationship with Steinerrules
The 1st and 2nd SteinerAxes are described in the Steiner rules (8) and (9). That’s why they are named after Steiner.
Rules 8, 9, 10 of Steiner about quadrilaterals say (see Ref4):
(8) Each of the four possible triangles in a quadrilateral has an incircle and three excircles. The centers of these 16 circles lie, four by four, on eight new circles.
(9) These eight new circles form two sets of four, each circle of one set being
orthogonal to each circle of the other set. The centers of the circles of each
set lie on a same line. These two lines are perpendicular.
(10) Finally, these last two lines intersect at the point F (Miquel Point).
Both sets of 4 circles as described in (9) are coaxal. One set has an axis with real intersection points of the 4 circles, the other set has an axis with imaginary intersection points of the 4 circles. These axes are called the 1^{st} Steineraxis (axis with 2 real common intersection points of 4 circles) and the 2^{nd} Steineraxis (axis with 2 imaginary common intersection points of 4 circles). The common intersection points on the 1^{st} Steiner Axis are the only points in the real plane that are invariant under ClawsonSchmidt Conjugation. See Ref15d. The Steiner Axes also can be constructed as the angle bisectors of QLP1.QLP4 and the axis of the inscribed Parabola QLCo1. See Ref34, Eckart Schmidt, QFG #914.
Construction:
Eckart Schmidt describes this construction in Ref15d:
 Choose one Quadrigon of the Reference Quadrilateral. For example S41.S12.S23.S34, where Sij = Intersection Li ^ Lj.
 Let P be a random point. Draw a circle Ci1 through P and the 2 vertices on a side L1 of the Quadrigon. Draw a circle Ci2 through P and the 2 vertices on the opposite side L3 of the Quadrigon. Let X be the 2^{nd} intersection point of Ci1 and Ci2.
 Draw a circle Ci3 through X and the 2 vertices on a third side L2 of the Quadrigon. Draw a circle Ci4 through X and the 2 vertices on the last side L4 of the Quadrigon. Let Y be the 2^{nd} intersection point of Ci3 and Ci4.
 Y is the ClawsonSchmidt Conjugate.
Coordinates:
Let Q (u : v : w) be a random point not on one of the defining lines of the Reference Quadrilateral.
The ClawsonSchmidt Conjugate in CTcoordinates gives this result:
(a^{2} m n (a^{2} (lm) ( l  n) v w + b^{2} (l  m) w ( l u +n v+n w) + c^{2} (l  n) v (l u+m v+m w)) :
b^{2} l n (b^{2} (ml) (mn) u w + c^{2} (m n) u (l u+m v+ l w) + a^{2} (m  l) w (n u+m v+n w)) :
c^{2} l m (c^{2} (n  l) (nm) u v + a^{2} (n – l) v (m u+m v+n w)+ b^{2} (n m) u (l u+ l v + n w)))
The ClawsonSchmidt Conjugate in DTcoordinates gives this result:
(b^{4} L N u (u + v  w) + c^{4} L M u (u  v + w)  a^{4} M N u (u + v + w)  2 b^{2} c^{2} L u (N v + M w)
+ 2 a^{2} b^{2} N ((u + v) (N^{2} u + M^{2} v)  N^{2} w^{2})  2 a^{2} c^{2} M (M^{2} v^{2} + (u + w) (M^{2} u + N^{2} w)) :
a^{4} M N v (u + v  w) + c^{4} L M v (u + v + w)  b^{4} L N v (u + v + w)  2 a^{2} c^{2} M v (N u + L w)
 2 a^{2} b^{2} N ((u + v) (L^{2} u + N^{2} v)  N^{2} w^{2}) + 2 b^{2} c^{2} L (L^{2} u^{2} + (v + w) (L^{2} v + N^{2} w)) :
 2 a^{2} b^{2} N (M u + L v) w + a^{4} M N w (u  v + w) + b^{4} L N w (u + v + w)  c^{4} L M w (u + v + w)
 2 b^{2} c^{2} L (L^{2} u^{2} + (v + w) (M^{2} v + L^{2} w)) + 2 a^{2} c^{2} M (M^{2} v^{2} + (u + w) (L^{2} u + M^{2} w)))
where:
L = m^{2}  n^{2} M = n^{2}  l^{2} N = l^{2}  m^{2}
Coordinates with Miquel Triangle as Reference Triangle
If we take the Miquel triangle as reference triangle ABC (C in the Miquel Point),
then the ClawsonSchmidt Conjugate QLTf1 gets a very simple form for Quadrigons:
(x : y : z) > (a^{2} y : b^{2} x : (a^{2} y z+b^{2} z x+c^{2} x y) / (x+y+z)).
Let L3^L4 = P4 = (u : v : w), then:
L2^L3 = P3 = (a^{2} w : T : c^{2} u),
L1^L2 = P2 = (a^{2} v : b^{2} u : T),
L4^L1 = P1 = (T : b^{2} w : c^{2} v)
where T = (a^{2} v w+b^{2} w u+c^{2} u v) / (u+v+w).
(x : y : z) > (a^{2} y : b^{2} x : (a^{2} y z+b^{2} z x+c^{2} x y) / (x+y+z)).
Let L3^L4 = P4 = (u : v : w), then:
L2^L3 = P3 = (a^{2} w : T : c^{2} u),
L1^L2 = P2 = (a^{2} v : b^{2} u : T),
L4^L1 = P1 = (T : b^{2} w : c^{2} v)
where T = (a^{2} v w+b^{2} w u+c^{2} u v) / (u+v+w).
See Ref34, QFG #338 by Eckart Schmidt.
Examples of ClawsonSchmidt Conjugates:
Point/Line/Curve 1

Point/Line/Curve 2



General


Defining Quadrilateral Lines: L1,L2,L3,L4

Circumcircle of corresponding QLComponent Triangle

Intersection point Li^Lj

Opposite intersection point Lk^Ll

A line not through QLP1

A circle through QLP1

A line through QLP1

A line through QLP1

A circle with QLP1 as center

Another circle with QLP1 as center



Points/Lines/Curves


QLP1: Miquel Point

Some (undefined) point at infinity

QLP4: Miquel Circumcenter


QLL2 Steiner Line

QLCi3 Miquel Circle

QLCo1: Inscribed Parabola

QLQu1: Morley’s Mono QLCardioide

Circle through X(186) points in QLCT’s
See QLP28, center of this circle.

Circle through X(265) points in QLCT’s
See QLP29, center of this circle.

QGP1: Diagonal Crosspoint

QAP4: Isogonal Center

QGP18: Quasi Isogonal Crosspoint


QGL1: 3^{rd} Diagonal

QGCi3: Quasi Isogonal Circle

QACi1: QACircumcircle DiagonalTriangle




Invariances


Steiner Bisector Circlei (i=18)
(circles in rule 9 of Steiner)

Same Steiner Bisector Circlei (i=18)

Schmidt Circle

Schmidt Circle

Intersection point
1^{st} SteinerAxis ^ Schmidt Circle

Same intersection point
1^{st} SteinerAxis ^ Schmidt Circle
(the only invariant points there are)

Intersection point
2^{nd} SteinerAxis ^ Schmidt Circle

Other intersection point
2^{nd} SteinerAxis ^ Schmidt Circle (switching points)

QLCu1: Quasi Isogonal Cubic

QLCu1: Quasi Isogonal Cubic

Performances in Quadrangles:
There are special surprises applying the ClawsonSchmidt Conjugate wrt Quadrangles:
 QLTf1 produces at Quadrigonlevel using QGP1 (the Diagonal Crosspoint) a QuadranglePoint: QAP4 (Isogonal Center) which is valid for all its QAQuadrigons. See Ref15d.
 Consider the 4 points of a QLQuadrigon as a QAQuadrigon also defining a Quadrangle. Let M2 and M3 be the Miquel Points of the other two QAQuadrigons. Now M2 and M3 are mutual ClawsonSchmidt Conjugates wrt the 1^{st} QAQuadrigon. See Ref15b and Ref15d.
 For a Quadrangle there are three ClawsonSchmidt Conjugates wrt the quadrilaterals
P1P2, P2P3, P3P4, P4P1 (CSCa),
P1P2, P2P4, P4P3, P3P1 (CSCb),
P1P4, P4P2, P2P3, P3P1 (CSCc).
Now the product of two of these conjugates achieves the third one.
The product of all three achieves the identity. See Ref15d. The QADTP4 Cubic (QACu1) of the quadrangle is invariant under these three transformations.
 Let P be some point on the QADTP4 Cubic (QACu1) and let Pa = CSCa(P), Pb = CSCb(P), Pc = CSCc(P). The tangents at P, Pa, Pb, Pc to the cubic QACu1 coincide in the Isogonal Center (QAP4) of the Quadrangle P.Pa.Pb.Pc. See Ref15b.
Properties:
 The intersections of lines L through P and the tangents T at the QLTf1imagecircle of L in QLP1 give an orthogonal hyperbola, centered in the midpoint of P.QLP1 with asymptotes parallel to the Steiner axes. See Ref34, Eckart Schmidt, QFGmessage #1666.