HYACINTHOS 29242

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

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

(Oa), (Ob), (Oc) = the circles (I, INa), (I, Nb), (I, INc), resp.

(O1), (O2), (O3) = the reflections of (Oa), (Ob), (Oc) in BC, CA, AB, resp.
The radical axes of the pairs of the circles are parallels to the bisectors AI, BI, CI.
Their intersection (radical center of the circles) lies on the OI line.

Point?


[César Lozada]:

 

Q = X(1)X(3) ∩ X(5)X(33337)

= a*(4*a^6-7*(b+c)*a^5-5*(b^2-4*b*c+c^2)*a^4+(b+c)*(14*b^2-25*b*c+14*c^2)*a^3-2*(b^4+c^4+b*c*(7*b^2-15*b*c+7*c^2))*a^2-(b^2-c^2)*(b-c)*(7*b^2-11*b*c+7*c^2)*a+3*(b^2-c^2)^2*(b-c)^2) : : (barys)

= 3*X(1)+X(26285)

= lies on these lines: {1, 3}, {5, 33337}, {1483, 10265}, {5901, 16174}, {10283, 11263}

= midpoint of X(i) and X(j) for these {i,j}: {1, 26287}, {1385, 11567}, {24680, 26086}

= X(26287)-of-anti-Aquila triangle

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1, 11849, 24680), (1, 21842, 13751), (1385, 24680, 36)

= [ 0.1706485290113946, 0.4354855380517134, 3.2604136344817800 ]

 

César Lozada