HYACINTHOS 29066

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle.

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.
Oa, Ob, Oc = the circumcenters of IBC, ICA, IAB, resp.
O1, O2, O3 = the reflections of Oa, Ob, Oc in BC, CA, AB, resp.
Ma, Mb, Mc = the midpoints of NaO1, NbO2, NcO3, resp.

ABC, MaMbMc are circumcyclologic.
ie the circumcircles of AMbMc, BMcMa, CMaMb and ABC are concurrent
the circmcircles of MaBC, MbCA, McAB and MaMbMc are concurrent

Cyclologic centers?


[Peter Moses]:

Hi Antreas,

 

(ABC, MaMbMc):  X(13624)X(28185)∩X(15178)X(28173)

= a*(a - b)*(a - c)*(3*a^3 + 13*a^2*b + 13*a*b^2 + 3*b^3 - 3*a^2*c - 7*a*b*c - 3*b^2*c - 3*a*c^2 - 3*b*c^2 + 3*c^3)*(3*a^3 - 3*a^2*b - 3*a*b^2 + 3*b^3 + 13*a^2*c - 7*a*b*c - 3*b^2*c + 13*a*c^2 - 3*b*c^2 + 3*c^3) : :

 

= lies on the circumcircle and these lines: {13624, 28185}, {15178, 28173}


(MaMbMc, ABC):

= (2*a + b + c)*(3*a^3 + a^2*b + a*b^2 + 3*b^3 + 9*a^2*c + 5*a*b*c + 9*b^2*c - 3*a*c^2 - 3*b*c^2 - 9*c^3)*(3*a^3 + 9*a^2*b - 3*a*b^2 - 9*b^3 + a^2*c + 5*a*b*c - 3*b^2*c + a*c^2 + 9*b*c^2 + 3*c^3) : :

Best regards,
Peter Moses.