The Cayley-Bacharach Point in an Octangle is defined as follows:
Let P1,...,P8 be eight distinct points in the plane, no four of which collinear, and no seven of them on a conic. There exists a unique ninth point P9 such that every cubic curve through P1,...,P8 also contains P9.
Let P1,...,P8 be eight distinct points in the plane, no four of which collinear, and no seven of them on a conic. There exists a unique ninth point P9 such that every cubic curve through P1,...,P8 also contains P9.
Many properties can be derived from the Cayley Bacharach Theorem, especially when a cubic is degenerated, for example in a conic and two lines, etc.
See also Ref-34, QFG-messages #732, #733, #736, #737, #1476, #1491, #1517, #1696, #1698-1701, #1733, #2431, #2439, #2440, #2442-2445, #2447-2450, #2452-2454, #2457-2471, #2477, #2480.
See also Ref-34, QFG-messages #732, #733, #736, #737, #1476, #1491, #1517, #1696, #1698-1701, #1733, #2431, #2439, #2440, #2442-2445, #2447-2450, #2452-2454, #2457-2471, #2477, #2480.
Calculation of coordinates:
In [69] is given a method for calculating the Cayley-Bacharach Point as follows:
Let P1 = (x1:y1:z1), P2 = (x2:y2:z2), etc.
Let C(P1,P2,P3,P4,P5,P6) = Determinant of
Let D(P1;P2,P3,P4,P5,P6,P7,P8) = Determinant of
Cx = C(P1,P4,P5,P6,P7,P8),
Cy = C(P2,P4,P5,P6,P7,P8),
Cz = C(P3,P4,P5,P6,P7,P8),
Dx = D(P1;P2,P3,P4,P5,P6,P7,P8),
Dy = D(P2;P3,P1,P4,P5,P6,P7,P8),
Dz = D(P3;P1,P2,P4,P5,P6,P7,P8).
The Cayley-Bacharach point CB is given by the formula:
CB = CxDyDz · P1 + DxCyDz · P2 + DxDyCz · P3
Correspondence with EQF:
Eckart Schmidt wrote next properties of the Cayley Bacharach Point (CB-point) in relationship to Quadrangles/Quadrigons at Ref-34, QFG#2640:
(1) Consider the vertices of a quadrangle and their complements wrt the remaining triangle.
The CB-point of these 8 points is QA-P1.
(2) Consider the vertices of a quadrangle and their anticomplements wrt the remaining triangle.
The CB-point of these 8 points is a new QA-point
... on a QA-circumscribed conic through QA-P10 and QA-P5
... and a parallel to QA-P1.QA-P16 through QA-P5,
... with tangent through QA-P16.
(3) Consider the vertices of a quadrangle and the in-/ex-centers of the Miquel triangle QA-Tr2.
The CB-point of these 8 points
... is the 3rd intersection of QA-Cu1 and Q.QA-P41
... with Q = intersection of QA-Cu1 and its asymptote (see EQF).
(4) Consider the vertices of a quadrangle and their isogonal conjugates wrt the Miquel triangle,
The CB-point of these 8 points is the tangential of Q wrt QA-Cu1.
(5) Consider the vertices of a quadrangle, its mid-DT and QA-P10.
The CB-point of these 8 points is the 2nd intersection of QA-P16.QA-P20 and the QA-circumconic through QA-P20.
(6) Consider the vertices of a quadrigon and the 4th parallelogram points of three consecutive vertices.
The CB-point of these 8 points is the reflection of QG-P1 in QG-P15.
(The QA-triangle of these CB-points is homothetic to QA-Tr1 wrt QA-P1 and factor -3.)
(7) Consider the vertices of QA-Tr1 and QA-Tr2 on QA-Cu1
... with Q and QA-P3: CB-point = QA-Tf2(res(QA-P4,QA-P41)) = QA-P41,
... with Q and QA-P4: CB-point = QA-Tf2(res(QA-P3,QA-P41)),
... with QA-P3 and QA-P4: CB-point = QA-Tf2(res(Q,QA-P41),
... with QA-P4 and QA-P41: CB-point = QA-Tf2(res(Q,QA-P3)).
Or: If you take the vertices of QA-Tr1 and QA-Tr2 and two of the points Q, QA-P3, QA-P4, QA-P41,
you get the CB-point as QA-Tf2-image of the 3rd intersection of QA-Cu1 and the line through the other two points.
(8) Consider the vertices of a quadrangle and Q, QA-P3, QA-P4, QA-P41.
The CB-point of these 8 points is res(Q,QA-Tf2(res(QA-P3,QA-P41))), where res(X, Y) = 3rd intersection of QA-Cu1 and XY.
Correspondence with EQF:
Eckart Schmidt wrote next properties of the Cayley Bacharach Point (CB-point) in relationship to Quadrangles/Quadrigons at Ref-34, QFG#2640:
(1) Consider the vertices of a quadrangle and their complements wrt the remaining triangle.
The CB-point of these 8 points is QA-P1.
(2) Consider the vertices of a quadrangle and their anticomplements wrt the remaining triangle.
The CB-point of these 8 points is a new QA-point
... on a QA-circumscribed conic through QA-P10 and QA-P5
... and a parallel to QA-P1.QA-P16 through QA-P5,
... with tangent through QA-P16.
(3) Consider the vertices of a quadrangle and the in-/ex-centers of the Miquel triangle QA-Tr2.
The CB-point of these 8 points
... is the 3rd intersection of QA-Cu1 and Q.QA-P41
... with Q = intersection of QA-Cu1 and its asymptote (see EQF).
(4) Consider the vertices of a quadrangle and their isogonal conjugates wrt the Miquel triangle,
The CB-point of these 8 points is the tangential of Q wrt QA-Cu1.
(5) Consider the vertices of a quadrangle, its mid-DT and QA-P10.
The CB-point of these 8 points is the 2nd intersection of QA-P16.QA-P20 and the QA-circumconic through QA-P20.
(6) Consider the vertices of a quadrigon and the 4th parallelogram points of three consecutive vertices.
The CB-point of these 8 points is the reflection of QG-P1 in QG-P15.
(The QA-triangle of these CB-points is homothetic to QA-Tr1 wrt QA-P1 and factor -3.)
(7) Consider the vertices of QA-Tr1 and QA-Tr2 on QA-Cu1
... with Q and QA-P3: CB-point = QA-Tf2(res(QA-P4,QA-P41)) = QA-P41,
... with Q and QA-P4: CB-point = QA-Tf2(res(QA-P3,QA-P41)),
... with QA-P3 and QA-P4: CB-point = QA-Tf2(res(Q,QA-P41),
... with QA-P4 and QA-P41: CB-point = QA-Tf2(res(Q,QA-P3)).
Or: If you take the vertices of QA-Tr1 and QA-Tr2 and two of the points Q, QA-P3, QA-P4, QA-P41,
you get the CB-point as QA-Tf2-image of the 3rd intersection of QA-Cu1 and the line through the other two points.
(8) Consider the vertices of a quadrangle and Q, QA-P3, QA-P4, QA-P41.
The CB-point of these 8 points is res(Q,QA-Tf2(res(QA-P3,QA-P41))), where res(X, Y) = 3rd intersection of QA-Cu1 and XY.