HYACINTHOS 29472

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle, P,P* two isogonal conjugate points, A'B'C' the pedal triangle of P
and La,Lb,Lc the parallels to PP* through A',B',C', resp.

 

Denote:

 

L1, L2, L3 =The reflections of La,Lb,Lc in the cevians AP*, BP*, CP*, resp. 

 

Then L1, L2, L3 are concurrent.

Which is the point of concurrence for some P's: O, H, G, K .... ?


Corollary.
For P = I, since I* = I, II* is any line passing through I.

So we have::

Let ABC be a triangle and A'B'C' the pedal triangle of I.

Let L be a line passing through I ("passing through I" is not necessary, because following we will take parallels to L)

Denote:

La, Lb, Lc = the parallels to L through A', B', C', resp.
L1, L2, L3 = the reflections of La, Lb, Lc in AI, BI, CI, resp.

L1, L2, L3 concur on the pedal circle of I (incircle)

Which is the point of concurrence for some lines: IO, IH, IG, ......... ?

[Peter Moses]:


Hi Antreas,

P = G ->  MIDPOINT OF X(12149) AND X(15534) =

 

a^2*(2*a^8*b^4 + 2*a^6*b^6 - 2*a^4*b^8 - 2*a^2*b^10 + 6*a^8*b^2*c^2 - 31*a^6*b^4*c^2 - 16*a^4*b^6*c^2 + 21*a^2*b^8*c^2 + 2*a^8*c^4 - 31*a^6*b^2*c^4 + 114*a^4*b^4*c^4 - 30*a^2*b^6*c^4 - 2*b^8*c^4 + 2*a^6*c^6 - 16*a^4*b^2*c^6 - 30*a^2*b^4*c^6 - 4*b^6*c^6 - 2*a^4*c^8 + 21*a^2*b^2*c^8 - 2*b^4*c^8 - 2*a^2*c^10) : :

 

= lies on these lines: {2, 2854}, {12149, 15534}
= midpoint of X(12149) and X(15534)

P = O -> X(1539).
P = H -> X(1511).

 

 

P = K -> X(111)X(15271)∩X(126)X(3258) =

 

= 2*a^8 - 5*a^6*b^2 + 2*a^4*b^4 - 3*a^2*b^6 - 5*a^6*c^2 + 2*a^4*b^2*c^2 + 8*a^2*b^4*c^2 + 4*b^6*c^2 + 2*a^4*c^4 + 8*a^2*b^2*c^4 - 16*b^4*c^4 - 3*a^2*c^6 + 4*b^2*c^6 : :

= X[11162] - 3 X[21358]

= lies on these lines: {111, 15271}, {126, 3258}, {141, 543}, {2793, 5026}, {3734, 33962}, {7761, 32424}, {7778, 10717}, {11162, 21358}

>Corollary.

OP{p,q,r} line -> a^2 (a-b-c) (-2 b c p+c (a-b+c) q+b (a+b-c) r)^2 : :

GI line -> X(6018)
OI line -> X(1317)
IH line -> X(1361)
IN line -> X(13756)
IK line -> X(3021)
IGe line -> X(1362)


IX(19) line -> REFLECTION OF X(1367) IN X(1) =

 

= (a - b - c)*(2*a^5 - a^3*b^2 - a^2*b^3 - a*b^4 + b^5 + a^2*b^2*c - b^4*c - a^3*c^2 + a^2*b*c^2 + 2*a*b^2*c^2 - a^2*c^3 - a*c^4 - b*c^4 + c^5)^2 : :

 

= lies on the incircle and these lines: {1, 1367}, {11, 4989}, {55, 26702}, {243, 3326}, {1358, 4292}, {1364, 10391}, {1365, 1836}, {1858, 3022}, {3057, 6020}, {5137, 14027}

= reflection of X(1367) in X (1)
= X(7)-Ceva conjugate of X (1375)
= crosspoint of X(7) and X (1375)


IX(21) line  -> X(34194)

Best regards,
Peter Moses.