7L-s-3P1: 7L-Conical Triplet Points

In a 7-Line there are 7 Component 6-Lines.
So 7 6L-s-Co1 conics can be constructed.
They mutually have 4 intersection points, three of which are common points 7L-s-3P1a/b/c.
However 2 of them can be imaginary points.
Finally 7L-s-3P1 is derived as follows:
• in a 5-Line there is the center of the inscribed conic 5L-s-P1,
• in a 6-Line the 6 versions of 5L-s-P1 lie on a conic 6L-s-Co1,
• in a 7-Line the 7 versions of 6L-s-Co1 have 3 common points 7L-s-3P1.
7L s 3P1 Conical Triplet 01


The determination of the 3 out of 4 points of intersection is not always easy.
Therefore these steps of construction:
• Construct Co6 (=6L-s-Co1 wrt Ref. 7-Line omitting L6) and Co7 (=6L-s-Co1 wrt Ref. 7-Line omitting L7). See 6L-s-Co1 for the way of construction.
• Because of their definition Co6 and Co7 will meet in the Center of the conic 5L-s-Co1 (L1,L2,L3,L4,L5). Therefore construct 5L-s-P1 (here called Ce) and it will lie on Co6 and Co7.
• Intersect Co6 and Co7 and draw their intersection points. Ce will be one of these points.
• When 2 points are imaginary the 2 other points will be real and one of them will be Ce. The other real point will be 7L-s-s3P1a, whereas 7L-s-3P1b and 7L-s-3P1c will be imaginary.
• When no points are imaginary construct the QA-Diagonal Triangle DT of the 4 intersection points. Now the vertices of the Anticevian Triangle of Ce wrt DT will be the 3 points 7L-s-3P1a, 7L-s-3P1b and 7L-s-3P1c.