5P-s-P3 5P-Quang Duong’s Point

Let P1.P2.P3.P4.P5 be a Pentangle (system of 5 random points).
Let Q1 = Midpoint(P1, QA-P4 wrt Quadrangle P2.P3.P4.P5).
Let Q2 = Midpoint(P2, QA-P4 wrt Quadrangle P3.P4.P5.P1).
Let Q3 = Midpoint(P3, QA-P4 wrt Quadrangle P4.P5.P1.P2).
Let Q4 = Midpoint(P4, QA-P4 wrt Quadrangle P5.P1.P2.P3).
Let Q5 = Midpoint(P5, QA-P4 wrt Quadrangle P1.P2.P3.P4).
Q1, Q2, Q3, Q4, Q5 are concyclic and 5P-s-P3 is the center of the circle through Q1, Q2, Q3, Q4, Q5.
This peculiar 5P-Point was discovered by Ngo Quang Duong.
See Ref-34, QFG-messages #2773, #2774, #2777 an #2780.

5P s P3 Quang Duong's Point 01

The coordinates are too complicated to mention here.


• The midpoints of 5P-s-P4 & 5P-s-P5 and of 5P-s-P5 & 5P-s-P6 lie on Quang Duong’s circle. See Ref-34, QFG-message #3792.
5P-s-P3 is the Nine-point Center X(5) of a 5P-triangle described by Eckart Schmidt in Ref-34, QFG-messages #3575, # 3579, #3581. Quang Duong’s circle is the Nine-point circle in this triangle. This 5P-triangle consists of the intersection points (unequal 5P-s-P4) of an orthogonal hyperbola and the 5P-circumscribed conic 5P-s-Co1, the orthogonal hyperbola being centered in the middle of 5P-s-P4.5P-s-P5, bearing 5P-s-P4 and 5P-s-P5 and with axes parallel to those of 5P-s-Co1.