HYACINTHOS 27061

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, A'B'C' the cevian triangle of H and P a point.

Denote:

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

Q = the Poncelet point of (ABCP) = the point of concurrence of (Na), (Nb), (Nc) and (N)

N1 = the other than Q intersection of the NPC (N) and QNa

N2 = the other than Q intersection of the NPC (N) and QNb

N3 = the other than Q intersection of the NPC (N) and QNc

1. P = O
A'N1, B'N2, C'N3 are concurrent.
The parallels to A'N1, B'N2, C'N3 through A, B, C, resp. concur at X(74)

2. P = I
ABC and Triangle bounded by (A'N1, B'N2, C'N3) are parallelogic.


[Peter Moses]:

Hi Antreas,

1). X(1986).

2).

(ABC,T) at: X(80).

(T,ABC) at:
a (a^2 b-b^3+a^2 c-2 a b c+b^2 c+b c^2-c^3) (a^3+b^3-a b c-b^2 c-b c^2+c^3)::
on lines {{65,2841},{113,1829},{119,517},{513,11570},{692,1718},{912,1878},{942,3937},{1393,7428},{1828,5446},{1845,8677},{2835,12736},{2840,5083},...}.
reflection of X(i) in X(j) for these {i,j}: {{942,3937},{13753,11715}}.
barycentric product X(10015)X(13589).
barycentric quotient X(13589)/X(13136).

Best regards,
Peter Moses.