Let ABC be a triangle, OaObOc the antipedal triangle of O and P a point.
Denote:
Pa, Pb, Pc = the reflections of P in BC, CA, AB, resp.
P1, P2, P3 = the reflections of P in ObOc, OcOa, OaOb resp.
Which is the locus of P such that the circumcircles of PPaP1, PPbP2, PPcP3 are coaxial?
The Euler line of ABC?
And which is the locus of the second point of the intersection of the circles as P moves on the Euler line?
[César Lozada]:
> Which is the locus of P such that the circumcircles of PPaP1, PPbP2, PPcP3 are coaxial?
The entire plane.
Interestingly, for any P the common radical axis is the trilinear polar of a point on the circum-conic with trilinear equation CyclicSum[ (S^2-SW*SA)*a*v*w ] = 0, through ETC’s 2, 110, 2396, 2421, 4230, 5649, 5968, 6331, 9513. This conic is the isogonal-conjugate of the line {6,523}=polar trilineal of X(98). Its center is:
Q* = complement of X(290)
= a^3*((b^2+c^2)*a^2-b^4-c^4)^2 : : (trilinears)
= On Steiner inellipse, cubic K357 and these lines: {2,290}, {3,1625}, {5,39}, {6,694}, {32,1147}, {99,10684}, {110,248}, {141,216}, {160,206}, {187,1511}, {230,3229}, {232,297}, {237,3289}, {542,5661}, {571,9233}, {574,3016}, {647,5642}, {684,2491}, {942,1015}, {1107,1146}, {1493,5007}, {1503,8841}, {2482,8552}, {2492,3163}, {3003,5181}, {3284,6593}, {5158,8542}, {10317,11597}
= midpoint of X(110) and X(9513)
= complement of X(290)
= [ 1.145074602850215, 0.99462086076225, 2.423584838525799 ]
> And which is the locus of the second point of the intersection P* of the circles as P moves on the Euler line?
Locus = line {110,237,2080,2698} = polar trilinear of {6,2966}/\{182,691}
If P is such that OP=t*OH, then P* = ((3*S^2-SW^2)*t+S^2+SW^2)*X( 2080)-t*(3*S^2-SW^2)*X(2698)
ETC pairs (P,P*): (3,2080), (23,110), (237,237)
Some others:
P*(G) = a*((b^4+3*b^2*c^2+c^4)*a^4-(b^ 2+c^2)*(b^4+c^4)*a^2-b^4*c^4) : : (trilinears)
= On Parry circle and lines: {2,51}, {22,1613}, {110,237}, {111,694}, {187,353}, {211,1078}, {323,7711}, {352,5106}, {512,9147}, {1495,9999}, {1627,1691}, {2421,5201}, {3229,9998}, {3849,6787}, {3978,4576} , {5162,8569}, {6310,6658}, {6784,8859}, {6786,7840}, {9156,11186}, {9208,9213}, {9301,11328}, {9855,9879}
= midpoint of X(9855) and X(9879)
= reflection of X(7840) in X(6786)
= [ -5.662695405446477, -4.55708104515845, 9.409118469530908 ]
P*(H) = ((b^4+b^2*c^2+c^4)*a^8-3*(b^2+ c^2)*(b^4+c^4)*a^6+(3*b^8+3*c^ 8+(b^4+b^2*c^2+c^4)*b^2*c^2)* a^4-(b^4-c^4)*(b^2-c^2)*(b^4+ c^4)*a^2-(b^2-c^2)^2*b^4*c^4)* a : : (trilinears)
= (-6*cos(2*A)+6*cos(4*A)+1)* cos(B-C)+(-2*cos(A)-2*cos(3*A) +2*cos(5*A))*cos(2*(B-C))-cos( 3*(B-C))+cos(5*A)-cos(A)+2* cos(3*A) : : (trilinears)
= On lines: {4,69}, {52,7785}, {98,2387}, {110,237}, {187,11464}, {217,1691}, {1216,2896}, {1298,3484}, {2548,3567}, {3331,5104}, {5890,9744}, {6241,8721}, {6785,7699}, {7800,7999}
= reflection of X(4) in X(5167)
= [ 3.940174522421549, 0.78690593314901, 1.277341363994350 ]
P*(N) = ((b^2+c^2)^2*a^8-3*(b^4+b^2*c^ 2+c^4)*(b^2+c^2)*a^6+3*(c^6+b^ 6)*(b^2+c^2)*a^4-(b^2+c^2)*(b^ 8+c^8-b^2*c^2*(2*b^4-b^2*c^2+ 2*c^4))*a^2-(b^2-c^2)^2*b^4*c^ 4)*a : : (trilinears)
= (-9*cos(2*A)+5*cos(4*A)+1/2)* cos(B-C)+(-2*cos(A)-cos(3*A)+ cos(5*A))*cos(2*(B-C))-1/2* cos(3*(B-C))+cos(5*A)-4*cos(A) +cos(3*A) :: (trilinears)
= On lines: {5,141}, {110,237}, {114,1154}, {143,1506}, {6243,7752}
= [ -3.261977923479470, -3.22108430058158, 7.376174193146769 ]
The point P’ on the Euler line such that P*(P’)= X(2698) is:
P’ = a^8-4*(b^2+c^2)*a^6+(3*b^4+b^ 2*c^2+3*c^4)*a^4+(b^2-c^2)^2* b^2*c^2 : : (barycentrics)
= 4*S^2*X(3)-(S^2+SW^2)*X(4)
= On lines: {2,3}, {6,7709}, {32,11257}, {74,6037}, {76,5171}, {98,187}, {99,511}, {114,316}, {182,3972}, {262,574}, {352,1499}, {385,2080}, {576,7757}, {1078,6248}, {1350,4048}, {1384,9755}, {1498,3360}, {1503,2076}, {2549,9753}, {2794,5162}, {3095,7783}, {3098,10000}, {3329,10796}, {3734,8722}, {3849,6054}, {5017,6776}, {5116,5480}, {5188,7816}, {5210,9756}, {7737,9744}, {7782,9737}, {7786,10358}, {7840,8724}, {8716,11477}, {8782,9301}, {8859,11632}
= reflection of X(i) in X(j) for these (i,j): (4,1513), (98,187), (316,114), (376,8598), (385,2080), (5999,3), (7464,7472), (7470,10997), (7840,8724), (8597,381)
= [ -13.807763809928720, -14.64085850325427, 20.149457511819830 ]
César Lozada
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