Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of N = X(5), N* = X(54), resp.
Denote:
Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp.
N'a, N'b, N'c = the reflections of Na, Nb, Nc in NA', NB', NC', resp.
N1, N2, N3 = the NPC centers of N*BC, N*CA, N*AB, resp.
N'1, N'2, N'3 = the reflections of N1, N2, N3 in N*A", N*B", N*C", resp.
Ma, Mb, Mc = the midpoints N'aN'1, N'bN'2, N'cN'3, resp.
1. ABC, N'aN'bN'c
2. ABC, N'1N'2N'3
3. ABC, MaMbMc
are orthologic.
APH
----------------------------------------------------------------------------------------
1) ABC->N’aN’bN’c = X(1154), N’aN’bN’c ->ABC = X(546)
2) ABC->N’1N’2N’3 = X(523), N’1N’2N’3 ->ABC = X(1493)
3) ABC->MaMbMc: Long and not interesting
MaMbMc->ABC =
= (19*R^2+6*SA-10*SW)*S^2+(23*R^2-8*SW)*SB*SC : : (barys)
= X(3)-3*X(8254), X(3)+3*X(20424), 5*X(5)-X(3519), 3*X(5)+X(15801), 3*X(54)+X(3627), 3*X(195)+5*X(3091), X(546)-3*X(3574), X(546)+3*X(22051), 5*X(546)-3*X(22804), 3*X(1209)-5*X(12812), X(1493)+3*X(3574), X(1493)-3*X(22051), 5*X(1493)+3*X(22804), 3*X(2888)-11*X(5072), 7*X(3090)-3*X(21230), X(3519)+5*X(11803), 3*X(3519)+5*X(15801), 5*X(3574)-X(22804), 3*X(11803)-X(15801), 5*X(22051)+X(22804)
= on lines: {3, 8254}, {4, 17507}, {5, 1173}, {30, 12242}, {54, 3627}, {113, 137}, {143, 12010}, {195, 3091}, {539, 3850}, {1154, 3628}, {1209, 12812}, {2888, 5072}, {2914, 11801}, {2918, 17714}, {3090, 21230}, {3518, 15806}, {3525, 12307}, {3851, 11271}, {3853, 10619}, {3857, 6288}, {5076, 12254}, {5079, 12316}, {6152, 13451}, {6689, 12108}, {7691, 14869}, {10272, 25714}, {10610, 12103}, {12102, 18400}, {12325, 15022}, {12606, 14449}, {12811, 20584}, {13432, 19709}, {15425, 16337}, {24144, 27423}
= midpoint of X(i) and X(j) for these {i,j}: {5, 11803}, {546, 1493}, {2914, 11801}, {3574, 22051}, {3853, 10619}, {8254, 20424}, {12606, 14449}
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (546, 22051, 1493), (1493, 3574, 546)
= [ -7.6173469302235330, 4.8797227079583880, 3.7780934980394340 ]
César Lozada
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου