HYACINTHOS 29571

[Tran Quang Hung]:

Let ABC be a triangle with circumcenter O.

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

Perpendiculars from A', B', C' to BC, CA, AB, respectively bound triangle A"B"C"..

Then the inverse of the orthocenter of A"B"C" in its circumcircle lies on the Euler line of ABC.

Which is this point?


[César Lozada]

 

Q = EULER LINE INTERCEPT OF X(8718)X(30210)

= (SB+SC)*(5*S^4-(72*R^4-2*(2*SA+23*SW)*R^2-5*SA^2+7*SA*SW+4*SB*SC+7*SW^2)*S^2+(4*R^2-SW)*(72*R^4-2*(8*SW+9*SA)*R^2-7*SB*SC+7*SA^2-4*SW^2)*SA) : : (barys)

= lies on these lines: {2, 3}, {8718, 30210}

= [ -26.8373623775178700, -27.6360846608290500, 35.1598134213358100 ]

 

César Lozada