HYACINTHOS 24022

 

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of H.

Denote:

Ab, Ac = the orthogonal projections of A' on HB, HC, resp

Abc, Acb = the midpoints of AbC', AcB', resp.

Similarly:

Bc, Ba = the orthogonal projections of B' on HC, HA, resp.

Bca, Bac = the midpoints of BcA', BaC', resp.

Ca, Cb = the orthogonal projections of C' on HA, HB, resp.

Cab, Cba = the midpoints of CaB', CbA', resp.

The six points Abc, Acb, Bca, Bac, Cab, Cba are concyclic.

Center of the circle?

Generalization:

For any point P instead of H, the six midpoints lie on a conic.

For which points (other than H) the conic is a circle?

 

APH

[Peter Moses]:

Hi Antreas,

 

>The six points Abc, Acb, Bca, Bac, Cab, Cba are concyclic.

>Center of the circle?

X(10110).

Squared radius (S^6 + SA^2 SB^2 SC^2) / (16 S^2 (S^2 SW - SA SB SC)).

 

Best regards,

Peter Moses.