HYACINTHOS 26344

[Le Viet An]:

 

Let ABC be a triangle and P a point.

Denote:

Ab, Ac = the orthogonal projections of A on PB, PC, resp.
Similarly Bc, Ba and Ca, Cb.

La, Lb, Lc = the Euler lines of AAbAc, BBcBa, CCaCb, resp.

A'B'C' = the triangle bounded by La, Lb, Lc.

For P = G, the centroid of A'B'C' lies on the NPC of ABC.

Which point is it?

 

[César Lozada]:

 

Q = reflection of X(11569) in X(5)

= (SB-SC)^2*(3*S^2-9*SB*SC-2*SW^ 2)*(3*S^2-(3*SA-2*SW)*(3*SA+2* SW)) : : (barycentrics)

= on the nine-points circle and these lines: {4,11568}, {5,11569}, {114,6032}, {126,9771}, {543,13234}, {2793,12494}, {3849,9127}, {6094,13377}

= midpoint of X(i) and X(j) for these {i,j}: {4,11568}, {6094,13377}

= reflection of X(11569) in X(5)

= antipode of X(11569) in the nine-points circle

= [ -0.786976228359526, 2.06720654330792, 2.572741288090982 ]

 

César Lozada