Τετάρτη 23 Οκτωβρίου 2019

HYACINTHOS 25418

[Tran Quang Hung]:

Let ABC be a triangle.

Denote:

A1B1C1 is pedal triangle of I

A', B', C' = the reflections of I in BC, CA, AB, resp.

A'', B'', C'' = the reflections of A1, B1, C1 in I, reps.

Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.

Na, Nb, Nc = the midpoints of AA'', BB'', CC'', resp.

The circumcircles of the trianges IMaNa, IMbNb, IMcNc are coaxial

2nd intersection?

 

[César Lozada]:

 

Q2 = Midpoint of X(11) and X(80)

= 2*a^4-2*(b+c)*a^3+(b^2+c^2)*a^ 2+2*(b^2-c^2)*(b-c)*a-3*(b^2- c^2)^2 : : (barycentrics)

= (R+r)*X(1)-3*r*X(5)

= on lines: {1,5}, {2,10609}, {4,653}, {8,4767}, {10,528}, {30,1155}, {44,5179}, {46,3627}, {65,546}, {100,405}, {104,3149}, {140,10572}, {149,1145}, {153,6835}, {214,6667}, {381,1159}, {382,1788}, {429,1862}, {497,5790}, {515,5126}, {517,11545}, {519,5087}, {632,3612}, {938,9654}, {942,2801}, {950,9956}, {960,2802}, {1086,6788}, {1320,3621}, {1478,4860}, {1479,5690}, {1482,10591}, {1656,3486}, {1698,6174}, {1728,5128}, {1770,3853}, {1836,3845}, {1985,3240}, {2646,3628}, {2771,7687}, {2800,6797}, {2829,6245}, {3035,3634}, {3245,3583}, {3295,5818}, {3419,3820}, {3474,3830}, {3485,3851}, {3526,4305}, {3579,5840}, {3586,10993}, {3622,10031}, {3625,5854}, {3654,9580}, {3679,4679}, {3843,4295}, {4187,5086}, {4304,11231}, {4663,5848}, {4870,11737}, {4997,6790}, {5204,10090}, {5217,10058}, {5220,5856}, {5225,6928}, {5229,5708}, {5550,6224}, {5560,10483}, {5657,9668}, {5691,11219}, {5714,9803}, {5855,11813}, {6147,10895}, {6914,11502}, {9779,11041}, {10246,10589}, {10573,10896}, {11604,11684}

= midpoint of X(i) and X(j) for these {i,j}: {11,80}, {149,1145}, {1317,9897}

= reflection of X(i) in X(j) for these (i,j): (214,6667), (1387,11), (3035,6702), (9945,3035)

= complement of X(10609)

= Fuhrmann circle-inverse-of-X(5722)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (355,9581,496), (1837,10826,5), (5587,5722,495), (7741,10950,5901)

= [ 3.119548427139289, 4.21397156327988, -0.716492028273672 ]

 

César Lozada

 

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