HYACINTHOS 27212

 [Antreas P. Hatzipolakis]:

 

The same property have the points O, H but for L = Euler line
Ie
Let ABC be a triangle, P = O or H, ApBpCp the pedal triangle of P and L = the Euler line.
Denote:
A', B', C'  = the orthogonal projections of A, B, C on L, resp.
The circles with diameters A'Ap, B'Bp, C'Cp and the circumcircle of ApBpCp (= NPC of ABC) are concurrent.

 

 

[César Lozada]:

For P=O, Q=point-of-concurrence=X(3258)

 

For P=H:

Q(H) = complement of X(10420)

= SB*SC*(S^2-3*SA^2)*(2*S^2+SA^ 2+2*SB*SC-SW^2)*(6*R^2-SA-SW) : : (barys)

= on the nine-point circle and on these lines: {2, 10420}, {4, 476}, {5, 12052}, {30, 131}, {113, 403}, {114, 468}, {115, 2501}, {122, 3154}, {125, 924}, {128, 186}, {133, 10151}, {136, 523}, {137, 2970}, {230, 1560}, {14120, 14672}

= complement of X(10420)

= Dou circles radical circle-inverse-of X(115)

= polar circle-inverse-of X(476)

= [ 2.8156522550690610, 3.3461241299764270, 0.0245851972765173 ]

 

Algebraic calculus shows that there other two points on the Euler line such that concurrence occurs, but they are real depending on the sign of the expression S^4+2*(18*R^2*(3*R^2-SW)+SW^2) *S^2-(4*R^2-SW)*SW^3 is positive. It seems to be always negative, but I could not prove it algebraically.

 

César Lozada