[César Lozada - Dao Thanh Oai]:
Theorem 17. Let ABC be a triangle P be a point in the plane and O its circumcenter. Denote Oa, Ob, Oc the centers of the circles (BCP), (CAP), (ABP). ObOc meets OA at A0 and define B0, C0 cyclically. Then A0Oa, B0Ob,C0Oc are concurrent. (Dao Thanh Oai)
For P=u:v:w (trilinears), the point of concurrence is:
Q(P) = a*((b*c*v*w+(u^2+v^2+w^2)*SA)*a*b*c+(S^2+SA*SB)*b*u*w+(S^2+SA*SC)*c*u*v) : :
Q(P) lies on the Brocard axis X(3)X(6)
If P lies on the Lemoine axis X(187)X(237) or in the line at infinity then Q(P)=X(511).
Let Q(P)=Q(t) be a point on the Brocard axis (not in the infinity) such that X(3)Q(t)=t*X(3)X(6). The locus of P such that Q(P)=Q(t) is a circle Γ{O*, ρ } with trilinear equation:
∑ [2*a*b*c*t*u^2+(2*t*(b^2+c^2)-a^2-b^2-c^2)*a*v*w] = 0
having center O*= Q(t) and squared-radius: ρ(t)2 = ((-3*tan(ω)^2+1)*t^2-2*t+1)*R^2
Therefore:
· Q(P) = Q(t=0)=X(3) for all the points P on the circle (X(3), ρ(t=0)=R), i.e, for P on the circumcircle of ABC.
· Q(P) = Q(t=1/2)=X(182) for all the points P on the circle (X(182), ρ(t=1/2)), i.e, for P on the Brocard circle of ABC, through ETC’s {3, 6, 1083, 1316, 5091, 5108, 6141, 6142, 6232, 6322, 6795, 8429, 9129, 11650, 13414, 13415, 13511, 13515, 13516, 14685, 18332, 18338, 22740, 22742, 24279}
· Q(P) = X(2080) ) for P on the circle centered at X(2080) through ETC’s { 23, 352, 385, 11676}
· Q(P) = X(9821) for P on the circle centered at X(11171) through ETC’s { 8782, 9862, 9984, 9998, 9999, 12499, 12503, 13210, 13235, 13236, 14691} (circumcircle of 5th Brocard triangle)
· Q(P) = X(11171) for P on the circle centered at X(11171) through ETC’s { 2, 353, 1340, 1341, 7709, 15920, 15921}
· Q(P) = X(13329) for P on the circle centered at X(13329) through ETC’s { 36, 238, 5526, 9441}
Others pairs (P,Q(P)):
(1, 991), (15, 15), (16, 16), (1742, 991), (2459, 6566), (2460, 6567), (7598, 1151), (7599, 1152), (7601, 6407), (7602, 6408), (7711, 12054), (8290, 12054), (10817, 6407), (10818, 6408), (10819, 1151), (11833, 6407), (11834, 6408), (11835, 1151), (11836, 1152), (14660, 12054), (14704, 5238), (14705, 5237)
Some others:
Q( X(4) ) = X(3)X(6) ∩ X(5)X(264)
= a^2*((b^4+b^2*c^2+c^4)*a^4-2*(b^4-c^4)*(b^2-c^2)*a^2+(b^4-b^2*c^2+c^4)*(b^2-c^2)^2)*(-a^2+b^2+c^2) : : (barys)
= (S^2-SB*SC)*(S^2+4*R^2*SW+2*SB*SC-SW^2) : : (barys)
= 3*X(2)-4*X(10003), 5*X(1656)-4*X(14767)
= on lines: {2, 1972}, {3, 6}, {4, 3164}, {5, 264}, {51, 6638}, {184, 13558}, {237, 6403}, {339, 7697}, {417, 15043}, {418, 3060}, {426, 5422}, {441, 18583}, {852, 5640}, {1073, 14489}, {1656, 14059}, {1942, 4846}, {1993, 6641}, {1994, 23606}, {5562, 17039}, {5889, 26897}, {5943, 6509}, {6375, 9243}, {6389, 14561}, {6776, 20975}, {10519, 20819}, {10796, 15013}, {11272, 28407}, {12161, 14152}, {15073, 20775}, {15526, 24206}, {20576, 28697}, {21969, 26907}
= midpoint of X(4) and X(3164)
= reflection of X(i) in X(j) for these (i,j): (3, 216), (264, 5)
= anticomplement of the anticomplement of X(10003)
= X(216)-of-X3-ABC reflections triangle
= X(264)-of-Johnson triangle
= X(3164)-of-Euler triangle
= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3, 5093, 15905), (3, 15851, 5050), (371, 372, 1970)
= [ -7.7535634177534960, -5.1664204548220370, 10.7959848357474100 ]
Q( X(5) ) = X(3)X(6) ∩ X(5012)X(14652)
= (SB+SC)*((5*R^2-2*SA-2*SW)*S^2-R^2*SA*SW) : : (barys)
= on lines: {3, 6}, {5012, 14652}
= [ 6.1281646724077430, 5.3971271669613320, -2.9241918671001170 ]
Q( X(13) ) = X(3)X(6) ∩ X(13)X(9159)
= (SB+SC)*(S^2+sqrt(3)*(SA-3*R^2+SW)*S+(9*R^2-SW)*SA) : : (barys)
= on lines: {3, 6}, {13, 9159}, {17, 10217}, {11658, 16241}
= [ 6.4331061155840160, 5.6291777759291700, -3.2255844932361300 ]
Q( X(14) ) = X(3)X(6) ∩ X(14)X(9159)
= (SB+SC)*(S^2-sqrt(3)*(SA-3*R^2+SW)*S+(9*R^2-SW)*SA) : : (barys)
= on lines: {3, 6}, {14, 9159}, {18, 10218}, {11659, 16242}
= [ 3.9527990040433850, 3.7417440126949400, -0.7741424518244484 ]
Q( X(20) ) = X(3)X(6) ∩ X(20)X(3186)
= (-a^2+b^2+c^2)*(2*(b^2+c^2)*a^6-(3*b^4-b^2*c^2+3*c^4)*a^4+2*(b^2+c^2)*b^2*c^2*a^2+(b^6-c^6)*(b^2-c^2))*a : : (barys)
= (S^2-SB*SC)*(S^2-16*R^2*SW-2*SB*SC+3*SW^2) : : (barys)
= on lines: {3, 6}, {20, 3186}, {185, 20794}, {2790, 23240}, {7750, 14615}, {9306, 14673}, {11328, 12294}
= midpoint of X(20) and X(3186)
= [ 9.9514788252431700, 8.3065459989827130, -6.7030114367312700 ]
César Lozada
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