Let ABC be a triangle and MaMbMc the cevian triangle of G = X(2).
Denote:
A', B', C' = the midpoints of AI, BI, CI, resp.
A", B", C" = the other than Ma, Mb, Mc intersections of the NPC and AG, BG, CG, resp.
The circumcircles of AA'A", BB'B", CC'C" are coaxial.
The one point of intersection lies on the NPC and is the Feuerbach point X(11).
The other one D lies on the circumcircle.
The common radical axis is the line X(2)X(11).
1. Which point is the point D?
2. Which point is the point D' = the other than D intersection of the circumcircle and the line X(2)X(11)?
3. Which point is the radical trace (intersection of the radical axis and the diacentric line)?
APH
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[César Lozada]
1) D=X(105)
2) D’=X(100)
3) D’’ = MIDPOINT OF X(11) AND X(105)
= 2*a^8-4*(b+c)*a^7+(b^2+10*b*c+c^2)*a^6+2*(b+c)*(b^2-4*b*c+c^2)*a^5-(3*b^4+3*c^4-(9*b^2-10*b*c+9*c^2)*b*c)*a^4+(b^2-c^2)*(b-c)*(4*b^2-13*b*c+4*c^2)*a^3-(b^4+c^4-(9*b^2+2*b*c+9*c^2)*b*c)*(b-c)^2*a^2-(b^2-c^2)*(b-c)*(2*b^4+2*c^4+3*(b-c)^2*b*c)*a+(b^4-c^4)*(b^2-c^2)*(b-c)^2 : : (barys)
= 3*X(11)-X(10773), 3*X(105)+X(10773), X(1292)-3*X(21154), X(5540)+3*X(16173), X(10743)-3*X(23513), X(20344)-5*X(31272)
= lies on these lines: {2, 11}, {676, 2826}, {952, 11716}, {1292, 21154}, {1387, 2809}, {2829, 5511}, {3756, 11219}, {5540, 16173}, {6713, 28915}, {10743, 23513}
= midpoint of X(11) and X(105)
= reflection of X(i) in X(j) for these (i,j): (120, 6667), (3035, 6714)
= [ 2.6685321540961740, 6.3611347441365740, -1.9948282582314830 ]
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