HYACINTHOS 29107

[Antreas P. Hatzipolakis]:

 

 

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

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

O' = the circumcenter of NaNbNc

(Oa), (Ob), (Oc) = the circumcircles of O'NbNc, O'NcNa, O'NaNb, resp.

 

(O1), (O2), (O3) = the reflections of (Oa), (Ob), (Oc) in IA', IB', IC', resp.

The radical center of (O1), (O2), (O3) lies on the OI line of ABC = Euler line of A'B'C'

Which point is it wrt triangles ABC and A'B'C'?

 

PS We may ask for the locus of P such that the radical center lies on the OP line of ABC or on the Euler line of A'B'C', but they are complicated, I think.

 

 

[César Lozada]:

 

 

The radical center (w/r to ABC) is:

 

O* = MIDPOINT OF X(3) AND X(11009)

= a*(2*a^6-4*(b+c)*a^5-2*(b^2-5*b*c+c^2)*a^4+(b+c)*(8*b^2-15*b*c+8*c^2)*a^3-2*(b^4+c^4+3*b*c*(b^2-3*b*c+c^2))*a^2-(b^2-c^2)*(b-c)*(4*b^2-7*b*c+4*c^2)*a+2*(b^2-c^2)^2*(b-c)^2) : : (barys)

= 3*X(1)+X(11014), X(35)-3*X(10246), 3*X(5790)-5*X(31262)

= lies on these lines: {1, 3}, {145, 26487}, {355, 5141}, {952, 25639}, {1125, 12619}, {1483, 3813}, {3622, 26492}, {5790, 31262}, {5836, 22935}, {5844, 31659}, {6265, 7504}, {11230, 30147}, {11231, 30144}, {17577, 21740}

= midpoint of X(i) and X(j) for these {i,j}: {3, 11009}, {1482, 11012}

= reflection of X(2646) in X(15178)

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 3, 11567), (1385, 10225, 13624), (10246, 32612, 1385)

= [ 3.4391451914956810, 3.1343686742939270, -0.1165808425248068 ]

 

César Lozada