HYACINTHOS 24984

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal triangles of I, resp.

Denote:

Oa, Ob, Oc = the circumcenters of IB'C', IC'A', IA'B', resp.

OaObOc, A"B"C" are homothetic.

Homothetic center?

Note: The Euler line of OaObOc is parallel to OI line (for the Euler line of A"B"C" is the OI line of ABC)

 

Locus:

Let ABC be a triangle and A'B'C', A"B"C" the cevian, pedal triangles of a point P, resp.

Denote:

Oa, Ob, Oc = the circumcenters of PB'C', PC'A', PA'B', resp.

Which is the locus of P such that OaObOc, A"B"C" are perspective?

 

[Peter Moses]:

Hi Antreas,

 

>OaObOc, A"B"C" are homothetic.

At a (a+b-c) (a-b+c) (b+c) (a^3-a b^2-3 a b c-2 b^2 c-a c^2-2 b c^2)::

On lines {{1,30},{12,3293},{37,65},{56, 4278},{226,2594},...}.

{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (1,79,500),(1,3649,1464).

crosssum of X(1) and X(500).

crossdifference of every pair of points on line X(3737) X(9404).

 

>Locus:

A nasty combination of infinity, degree 5 and degree 8.

 

Best regards,

Peter Moses.