HYACINTHOS 25106

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle,  A'B'C' the cevian triangle of O and A"B"C" the pedal triangle of O wrt triangle A'B'C'.

The circles AOA", BOB", COC" are coaxial.

Second point of intersection?

Generalization:

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

Which is the locus of P such that the circumcircles of APA", BPB", CPC" are coaxial?

O, I, H lie on the locus.

The Darboux cubic?

 

[Peter Moses]:

Hi Antreas,

 

Yes, Darboux cubic K004.

 

P = I, second = X(1155).

 

P = O, second  = {{3,64},{20,6526},{30,133},{ 186,1301},{250,2071},{253, 10304},{376,459},{1304,5896},{ 1559,6716}}.

isog X(10152).

reflection of X(1559) in X(6716).

midpoint of X(186) & X(2693).

on Q071,K039,K446.

inverse of X(64) in the circumcircle.

.....

 

P= H, second = X(5203).

 

Best regards,

Peter Moses.