HYCINTHOS 26322

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle and P a point.

Denote:

P* = the Poncelet point of (ABCP)
(ie the point the NPCs of ABC, PBC, PCA, PAB concur at)

A*, B*, C* = the midpoints of AP, BP, CP.

A1 = the other than P* intersection of P*A* and the NPC of PBC
B1 = the other than P* intersection of P*B* and the NPC of PCA
C1 = the other than P* intersection of P*C* and the NPC of PAB

A2 = the antipode of A1 in the NPC of PBC
B2 = the antipode of B1 in the NPC of PCA
C2 = the antipode of C1 in the NPC of PAB

P*, A2, B2, C2 are concyclic.

Which is the center of the circle for P = 

 

1. I
2. O
3. N

?

[Peter Moses]:

Hi Antreas,

 

Center of the circle for P{p,q,r} is:

p (c^2 (a^2-b^2-c^2) p^2 q^2+c^2 (a^2-b^2+c^2) p q^3-2 (a^4-2 a^2 b^2+b^4-2 a^2 c^2+c^4) p^2 q r-(2 a^4-4 a^2 b^2+2 b^4-3 a^2 c^2-3 b^2 c^2+c^4) p q^2 r+2 a^2 c^2 q^3 r+b^2 (a^2-b^2-c^2) p^2 r^2-(2 a^4-3 a^2 b^2+b^4-4 a^2 c^2-3 b^2 c^2+2 c^4) p q r^2-2 a^2 (a^2-b^2-c^2) q^2 r^2+b^2 (a^2+b^2-c^2) p r^3+2 a^2 b^2 q r^3)::

 

1). X(1).

 

2). X(12038).

 

3). (a^2 b^2-b^4+a^2 c^2+2 b^2 c^2-c^4) (a^12-6 a^10 b^2+14 a^8 b^4-16 a^6 b^6+9 a^4 b^8-2 a^2 b^10-6 a^10 c^2+18 a^8 b^2 c^2-14 a^6 b^4 c^2-3 a^4 b^6 c^2+6 a^2 b^8 c^2-b^10 c^2+14 a^8 c^4-14 a^6 b^2 c^4-3 a^4 b^4 c^4-4 a^2 b^6 c^4+4 b^8 c^4-16 a^6 c^6-3 a^4 b^2 c^6-4 a^2 b^4 c^6-6 b^6 c^6+9 a^4 c^8+6 a^2 b^2 c^8+4 b^4 c^8-2 a^2 c^10-b^2 c^10):: 

on lines {{2,11016},{5,128},{140,389},{ 195,252},{930,7604},{1487, 1656}}.

 

4). P -> X(n).

{{1,1},{2,12040},{3,12038},{4, 5},{6,12039},{13,5459},{14, 5460},{74,12041},{80,11},{98, 12042},{110,1511},...}.

 

Best regards,

Peter Moses.