Given a random point P.
QA-Tf15 is the center of the circumscribed conic through P and the vertices of the Reference Quadrangle.
QA-Tf15 is the center of the circumscribed conic through P and the vertices of the Reference Quadrangle.
Some construction:
Let P1.P2.P3.P4 be the Reference Quadrangle.
Let P be some random point.
1. Construct the circumscribed circles of the P-Cevian Triangles of the 4 QA-Component Triangles P1.P2.P3, P2.P3.P4, P3.P4.P1, P4.P1.P2.
2. There are 6 radical axes of these 4 circles and they coincide in one point !
3. This point as QA-Tf15(P).
See Ref-34, QFG-message #3032.
CT-Coordinates of QA-Tf15[(x:y:z)]:
( p (q - r)2 (q + r) (2 p + q + r) (p2 + p q + q2 + p r + 3 q r + r2) :
q (p - r)2 (p + r) (p + 2 q + r) (p2 + p q + q2 + 3 p r + q r + r2) :
r (p - q)2 (p + q) (p + q + 2 r) (p2 + 3 p q + q2 + p r + q r + r2) )
( p (q - r)2 (q + r) (2 p + q + r) (p2 + p q + q2 + p r + 3 q r + r2) :
q (p - r)2 (p + r) (p + 2 q + r) (p2 + p q + q2 + 3 p r + q r + r2) :
r (p - q)2 (p + q) (p + q + 2 r) (p2 + 3 p q + q2 + p r + q r + r2) )
Properties:
• All points QA-Tf15(P) lie on the Nine-point Conic QA-Co1. See Ref-34, QFG-message #3032.
• QA-Tf15(QA-P1) = QA-Tf15(QA-P16) = QA-Tf2(InfinityPoint of the line through the 3 QA-versions of QL-P18). See Ref-34, QFG-message #3032.
• QA-Tf15(QA-P5) = QA-Tf15(QA-P10). See Ref-34, QFG-message #3032.
• All points on QA-Co2 are mapped by QA-Tf15 into QA-P2. See Ref-34, QFG-message #3038. Likewise all points on some QA-circumscribed Conic will be mapped into its Conic Center.
• All points QA-Tf15(P) lie on the Nine-point Conic QA-Co1. See Ref-34, QFG-message #3032.
• QA-Tf15(QA-P1) = QA-Tf15(QA-P16) = QA-Tf2(InfinityPoint of the line through the 3 QA-versions of QL-P18). See Ref-34, QFG-message #3032.
• QA-Tf15(QA-P5) = QA-Tf15(QA-P10). See Ref-34, QFG-message #3032.
• All points on QA-Co2 are mapped by QA-Tf15 into QA-P2. See Ref-34, QFG-message #3038. Likewise all points on some QA-circumscribed Conic will be mapped into its Conic Center.