Σάββατο 2 Νοεμβρίου 2019

HYACINTHOS 29647

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, A'B'C' the pedal triangle of I and (Ia), (Ib), (Ic) the excircles.

Denote:

(Na), (Nb), (Nc) = the NPCs of IBC, ICA, IAB, resp.

Ra = the radical axis of (Na), (Ia)
Rb = the radical axis of (Nb), (Ib)
Rc = the radical axisof (Nc), (Ic)

A*B*C* = the triangle bounded by Ra, Rb, Rc

1. A'B'C', A*B*C*
2. IaIbIc, A*B*C*
3. NaNbNc, A*B*C*
are orthologic.

Orthologic centers?


[César Lozada]:

 

 

1)

A*B*C* à A’B’C’ = MIDPOINT OF X(9782) AND X(32635)

= (b^2+6*b*c+c^2)*a^2+12*(b+c)*b*c*a-(b^2-c^2)^2 : : (barys)

= 8*X(5)-3*X(13865), 5*X(1698)-X(5506), X(5557)+9*X(19875), 7*X(9780)+X(9782), 7*X(9780)-X(32635), 3*X(10172)-X(34198)

= lies on these lines: {2, 3303}, {5, 40}, {7, 12}, {10, 354}, {11, 3634}, {20, 26040}, {55, 17552}, {142, 3983}, {442, 3828}, {443, 9657}, {474, 31157}, {495, 11034}, {496, 19872}, {528, 17536}, {548, 5251}, {631, 4413}, {993, 17583}, {1210, 12620}, {1329, 18231}, {1574, 31462}, {2550, 9670}, {2551, 9656}, {2886, 19877}, {3058, 16842}, {3214, 17245}, {3526, 19854}, {3614, 3841}, {3649, 3740}, {3698, 11362}, {4208, 31141}, {4301, 25917}, {4309, 11108}, {4317, 9708}, {4421, 31259}, {4866, 5557}, {4999, 9342}, {5067, 31245}, {5084, 9671}, {5259, 6154}, {5260, 15326}, {5298, 17531}, {5657, 7958}, {5787, 10857}, {6057, 28612}, {6067, 7080}, {6174, 6675}, {7173, 33108}, {7486, 31246}, {8582, 10177}, {8715, 17590}, {9709, 31452}, {9844, 10395}, {9940, 12619}, {9956, 22798}, {10172, 34198}, {12616, 12671}, {12623, 12866}, {16239, 31235}, {16408, 31494}, {17527, 19876}, {17559, 31140}

= midpoint of X(9782) and X(32635)

= X(1173)-of-4th Euler triangle

= {X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (10, 17529, 15888), (3826, 9711, 4197), (3826, 9780, 12), (4197, 9711, 12), (4197, 9780, 9711), (4413, 19855, 24953), (5084, 31420, 9671), (8728, 19875, 21031)

= [ 2.7503993823312540, 1.7039833043620100, 1.1915686331962550 ]

 

A’B’C’ à A*B*C* = X(1)X(550) ∩ X(7)X(12)

= (2*a+b+c)*(2*a+3*b+3*c)*(a-b+c)*(a+b-c) : : (barys)

= X(1)-3*X(5557), 7*X(9780)-15*X(9782), 7*X(9780)-5*X(32635), 3*X(9782)-X(32635)

= lies on these lines: {1, 550}, {7, 12}, {11, 11544}, {57, 5506}, {65, 3625}, {145, 5434}, {553, 1125}, {1071, 31673}, {1317, 4298}, {3337, 3652}, {3634, 3982}, {3679, 5586}, {4031, 19862}, {4355, 10944}, {4860, 7965}, {5433, 21454}, {5708, 7173}, {6797, 11570}, {9776, 28647}, {11495, 30340}, {11551, 13624}, {11684, 26842}, {12690, 33667}, {14100, 15009}, {16006, 18483}, {17718, 31425}, {23958, 31260}

= X(1173)-of-intouch triangle

= X(2889)-of-inverse-in-incircle triangle

= [ 0.6885865080000078, 0.7685580933936743, 2.7907766443272150 ]

 

2)

A*B*C* à IaIbIc = A*B*C* à A’B’C’

IaIbIc à A*B*C* = X(5506)

 

3)

A*B*C* à NaNbNc = X(11)X(3634) ∩ X(119)X(12811)

= (b^2-10*b*c+c^2)*a^5-(b+c)^3*a^4-(2*b^4+2*c^4+(7*b^2-58*b*c+7*c^2)*b*c)*a^3+2*(b+c)*(b^4-8*b^2*c^2+c^4)*a^2+(b^2+17*b*c+c^2)*(b^2-c^2)^2*a-(b^2-c^2)^3*(b-c) : : (barys)

= lies on these lines: {11, 3634}, {119, 12811}, {1156, 5852}, {1484, 5535}

= [ 2.5358708601591540, 5.0415177863158530, -1.0200189979231290 ]

 

NaNbNc à A*B*C* = X(5506)

 

 

César Lozada

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