HYACINTHOS 25728

 

[Tran Quang Hung]:

 

Let ABC be a triangle.

A'B'C' is cevian triangle of circumcenter O.

A'',B'',C'' are midpoints of AA',BB',CC', resp..

The isogonal conjugate of O wrt triangle A''B''C'' lies on Euler line.

 

Which is this point ?

 

 

[César Lozada]:

Z = 2*(3*R^2-SW)*S^2-(4*R^2-SW)*(S B+SC)*SA : :  (barycentrics)

= 5*X(3091)-3*X(7565)

= Shinagawa coefficients: (E+2*F, 2*F)

= nine-points circle-inverse-of-X(3153)

= On lines: {2,3}, {12,9630}, {68,7592}, {125,9729}, {141,11444}, {343,5889}, {389,3580}, {511,3574}, {567,12370}, {569,9927}, {1176,1503}, {1181,11442}, {1209,11802}, {1352,11441}, {1568,11793}, {1614,12134}, {2888,3564}, {4348,7741}, {5012,6146}, {5448,5891}, {5449,9730}, {5890,12359}, {6800,9833}, {7221,7951}, {7699,7999}, {9019,11743}, {11402,12429}

= midpoint of X(4) and X(7512)

= reflection of X(i) in X(j) for these (i,j): (3,7568), (5576,5)

= orthocentroidal circle-inverse-of-X(7503)

= {X(i),X(j)}-Harmonic conjugate of X(k) for these (i,j,k): (2,4,7503), (2,3091,6816), (3,4,12225), (4,5,5133), (4,3090,7404), (4,3091,7566), (4,3547,22), (4,7494,20), (4,7558,3), (4,12088,7553), (5,235,3091), (5,10024,403), (5,11563,3850), (381,7387,4), (546,7553,4), (1656,10254,5), (3146,5169,1595), (3542,7401,1995), (3575,6676,7488), (3832,7500,4), (5133,7495,858)

= [ 31.706902165289140, 30.75373863389066, -32.284340187073370 ]

 

César Lozada