Dimidium-associated QLs
1. Introduction
 Let’s call QL the complete quadrilateral, VR the Van Rees circular isogonal cubic QLCu1 and Cl-S the Clawson-Schmidt transformation QL-Tf1
1 QL determines at the same time 1 VR and 1 Cl-S
1 VR determines 1 Cl-S
VR is given by P1, the Newton Line, the Steiner axes and the Schmidt circle
Cl-S is given by P1, the Steiner axes and the Schmidt circle
For a VR, there are an infinity of QLs and for a Cl-S, there are an infinity of VR and an
infinity2 of QL’s
 In order to have a QL, it’s VR and it’s Cl-S, we need 5 elements, for example a
reference triangle and a line or the diagonal triangle and the Newton Line or 5
independant points
2. Converse construction
 If we have P1, P4, P5, we have 3 independant points with the Miquel and Plücker
circles.
The 2 circles cut in P1 and another point, which is the reflexion of P1 in P4P5.
If we have the conjugate of this point, which is CSCe(P4P5), this gives the Cl-S and we
are able to find the Steiner axes and the Schmidt circle with the points F1 and F2.
The perpendicular bisector of the segment joining P1 to CSC(P4) is the Steiner Line,
which cuts the Plücker circle in the Plücker points and the perpendicular bisector of
the segment joining P1 to CSC(P5) is the CSC of the Plücker circle and cuts the Miquel
circle in 2 points, which are the CSC X and Y of the Plücker points.
 Now the choice of the Newton Line gives the 5th element and at the same time the
VR and the QL :
o If we choose the ordinary Newton Line, which is the perpendicular bisector
of the segment joining the Plücker points, we find the original QL with P4 as
Miquel point and P5 as Clawson center
o If we swap now the points P4 and P5 and the Miquel and Plücker circles, we
swap the same way the Plücker points and their CSC X and Y and have as
Newton Line the perpendicular bisector of XY and as Steiner Line the line XY.
3. Properties of the 2 QLs
 The 2 QLs share
o The Cl-S with P1, the Steiner axes and the Schmidt circle
o The Dimidium point P6 and the Dimidium circle, the rectangular hyperbola
with center P6 through P4 and P5 and axes parallel to the Steiner axes.
o The point P17 and therefore the 2nd conic Co (message 481) with the quartic
(messages 364 and 465), which passes through the vertices of the 2QLs and
through the Plücker points and their CSC X and Y.
It’s easy to check that the reflexion of P17 in the Newton Line is on the line L6 for the 2 QLs


The 2 QLs have symmetrical properties
o The pedal quadrangles of X and Y wrt the 1rst QL are orthocentric ; it’s the
same for the Plücker points wrt the associated QL
o The 1rst conic Co described in message 471 is inscribed in the DT of the 1rst
QL and tangent to XY as perpendicular bisector of P1CSC(P5) ; the same conic
inscribed in the DT of the associated QL is tangent to the initial Steiner Line,
as perpendicular bisector of the initial P1CSC(P4), becoming precisely
P1CSC(P5) of the associated QL.
The 2 diagonal triangles of the QL’s share the same Miquel point P17 and the same
Cl-S transformation, the vertices of the DT being the CSC-conjugates of the middles
of the diagonals and the circle circumscribed to the DT being the CSC-conjugate of
the Newton Lines for the 2 QLs.
The figure is terribly complicate, but I hope the idea is clear : the 2 QLs have vertices AA’, BB’
and CC’ and DD’, EE’ and FF’, the 2 DT have vertices α, β and γ and λ, μ and ν.
Despite all my efforts, I didn’t find an easy construction, so I used the search of 2 points A
and A’ (the same for D and D’) such as the circle with diameter AA’ and center on the Newton
Line passes through the Plücker points P1a and b, and there are one circle centered on the
2nd Steiner axis through A and A’ and F1 and F2 and it’s easy to check that a 3rd circle
centered on the other Newton Line passes through A and A’ and the points Q1a and Q1b.