Κυριακή 27 Οκτωβρίου 2019

HYACINTHOS 28378

[Antreas P. Hatzipolakis]:
 
 
Let ABC be a triangle and P a point.
 
Denote:
 
Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.
 
A', B', C' = the orthogonal projections of Na, Nb, Nc on BC, CA, AB, resp.
 
R1, R2, R3 = the reflections of NaA', NbB', NcC' in NbNc, NcNa, NaNb, resp.
 
A*B*C* = the triangle bounded by R1, R2, R3
 
For P = N:
 
ABC, A*B*C* are parallelogic.

Parallelogic centers?


[Peter Moses]:
 
Hi Antreas,

Parallelogic center (ABC, A*B*C*):
 
=  X(3)X(252)∩X(54)X(1263)

Barycentrics (a^4-2 a^2 b^2+b^4-a^2 c^2-b^2 c^2) (a^4-a^2 b^2+b^4-2 a^2 c^2-2 b^2 c^2+c^4) (a^4-2 a^2 b^2+b^4-a^2 c^2-2 b^2 c^2+c^4) (a^4-a^2 b^2-2 a^2 c^2-b^2 c^2+c^4) (a^6-a^4 b^2-a^2 b^4+b^6+2 a^2 b^3 c-2 b^5 c-a^4 c^2-a^2 b^2 c^2-b^4 c^2+2 a^2 b c^3+4 b^3 c^3-a^2 c^4-b^2 c^4-2 b c^5+c^6) (a^6-a^4 b^2-a^2 b^4+b^6-2 a^2 b^3 c+2 b^5 c-a^4 c^2-a^2 b^2 c^2-b^4 c^2-2 a^2 b c^3-4 b^3 c^3-a^2 c^4-b^2 c^4+2 b c^5+c^6) : : 
= lies on the cubic K466 and these lines: {3,252},{54,1263},{12026,19268},{14072,21975}.
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (930, 1141, 252), (1263, 6343, 10285).
= circumcircle inverse of X(252).

This point is the product of X(252) with
 
P1 = 
 
= X(2)X(2006)∩X(81)X(1086)
 
Barycentrics a^6-a^4 b^2-a^2 b^4+b^6+2 a^2 b^3 c-2 b^5 c-a^4 c^2-a^2 b^2 c^2-b^4 c^2+2 a^2 b c^3+4 b^3 c^3-a^2 c^4-b^2 c^4-2 b c^5+c^6 : : 
= lies on these lines: {2,2006},{81,1086},{278,23958},{651,1994},{3960,4560},{5718,17395},{5723,17484},{17013,19785},{17732,24046}.,

and P2 =
 
= X(2)X(2006)∩X(343)X(17483) 
 
Barycentrics a^6-a^4 b^2-a^2 b^4+b^6-2 a^2 b^3 c+2 b^5 c-a^4 c^2-a^2 b^2 c^2-b^4 c^2-2 a^2 b c^3-4 b^3 c^3-a^2 c^4-b^2 c^4+2 b c^5+c^6 : : 
= lies on these lines: {2,2006},{343,17483},{3448,21028}.

P1 P2 on lines {{2,94},{648,14129},{1994,11538}}.


Parallelogic center (A*B*C*, ABC) :
 
 = MIDPOINT OF X(1157) AND X(19553)
 
Barycentrics (2 a^4-3 a^2 b^2+b^4-3 a^2 c^2-2 b^2 c^2+c^4) (a^6-2 a^4 b^2+a^2 b^4+a^2 b^3 c-b^5 c-2 a^4 c^2-a^2 b^2 c^2+a^2 b c^3+2 b^3 c^3+a^2 c^4-b c^5) (a^6-2 a^4 b^2+a^2 b^4-a^2 b^3 c+b^5 c-2 a^4 c^2-a^2 b^2 c^2-a^2 b c^3-2 b^3 c^3+a^2 c^4+b c^5) : : 
= lies on these lines: {140,1493},{195,252},{1157,19553},{5965,10615},{7159,14102}.
= midpoint of X(1157) and X(19553).
= {X(195),X(252)}-harmonic conjugate of X(13856).

This point is the product of X(140) with,
 
= X(239)X(514)∩X(323)X(4858) 
 
Barycentrics (a^6-2 a^4 b^2+a^2 b^4+a^2 b^3 c-b^5 c-2 a^4 c^2-a^2 b^2 c^2+a^2 b c^3+2 b^3 c^3+a^2 c^4-b c^5) : : 
= lies on these lines: {239,514},{323,4858},{1994,14213},{3187,17035},{3219,18662},{17479,19742}.

and
 
= X(2)X(7)∩X(37)X(16698) 
 
Barycentrics (a^6-2 a^4 b^2+a^2 b^4-a^2 b^3 c+b^5 c-2 a^4 c^2-a^2 b^2 c^2-a^2 b c^3-2 b^3 c^3+a^2 c^4+b c^5) : : 
= lies on these lines:  {2,7},{37,16698},{81,18359},{323,6358},{1994,14213},{17168,22002} .

Best regards,
Peter Moses.
 

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου