5P-s-P2: 5P-Involutary Center

Let P1.P2.P3.P4.P5 be a Pentangle (system of 5 independent random points).
          Let Q1 = Involutary Conjugate (QA-Tf2) of P1 wrt Quadrangle P2.P3.P4.P5.
          Let Q2 = Involutary Conjugate (QA-Tf2) of P2 wrt Quadrangle P3.P4.P5.P1.
          Let Q3 = Involutary Conjugate (QA-Tf2) of P3 wrt Quadrangle P4.P5.P1.P2.
          Let Q4 = Involutary Conjugate (QA-Tf2) of P4 wrt Quadrangle P5.P1.P2.P3.
          Let Q5 = Involutary Conjugate (QA-Tf2) of P5 wrt Quadrangle P1.P2.P3.P4.
          Let R1 = Involutary Conjugate (QA-Tf2) of Q1 wrt Quadrangle Q2.Q3.Q4.Q5.
          Let R2 = Involutary Conjugate (QA-Tf2) of Q2 wrt Quadrangle Q3.Q4.Q5.Q1.
          Let R3 = Involutary Conjugate (QA-Tf2) of Q3 wrt Quadrangle Q4.Q5.Q1.Q2.
          Let R4 = Involutary Conjugate (QA-Tf2) of Q4 wrt Quadrangle Q5.Q1.Q2.Q3.
          Let R5 = Involutary Conjugate (QA-Tf2) of Q5 wrt Quadrangle Q1.Q2.Q3.Q4.
Now Pentangle P1.P2.P3.P4.P5 is point perspective with R1.R2.R3.R4.R5.
The perspector 5P-s-P2 is a regular Pentangle Center.
5P-s-P2 is the dual case of 5L-s-L2.
See Ref-34, QFG-messages #704 and #784.
5P s P2 Involutary Center 01

When using barycentric coordinates: P1=(1:0:0), P2=(0:1:0), P3=(0:0:1), P4=(p:q:r), P5=(x:y:z),
then 5P-s-P2 has coordinates:
 ((-r y + q z) (p2 q2 r6 x6 y4 - p3 q r6 x5 y5 + 2 q5 r5 x8 y z - 5 p q4 r5 x7 y2 z - 2 p2 q3 r5 x6 y3 z + 5 p3 q2 r5 x5 y4 z + 2 p4 q r5 x4 y5 z - p5 r5 x3 y6 z - 5 p q5 r4 x7 y z2 + 15 p2 q4 r4 x6 y2 z2 - 20 p4 q2 r4 x4 y4 z2 + 5 p5 q r4 x3 y5 z2 + p6 r4 x2 y6 z2 - 2 p2 q5 r3 x6 y z3 + 10 p4 q3 r3 x4 y3 z3 - 2 p6 q r3 x2 y5 z3 + p2 q6 r2 x6 z4 + 5 p3 q5 r2 x5 y z4 - 20 p4 q4 r2 x4 y2 z4 + 15 p6 q2 r2 x2 y4 z4 - 5 p7 q r2 x y5 z4 - p3 q6 r x5 z5 + 2 p4 q5 r x4 y z5 + 5 p5 q4 r x3 y2 z5 - 2 p6 q3 r x2 y3 z5 - 5 p7 q2 r x y4 z5 + 2 p8 q r y5 z5 - p5 q5 x3 y z6 + p6 q4 x2 y2 z6 : : )

5P-s-P2 lies on a conic mentioned at 5P-s-L1.