## Παρασκευή, 24 Φεβρουαρίου 2012

### Oa,Ob,Oc

Let ABC be a triangle and A'B'C' the cevian triangle of I.

The reflection of BB' in AA' intersects AB in Ab
The reflection of CC' in AA' intersects AC in Ac

Let Oa be the circumcenter of A'AbAc

A'AbAc is the reflection of A'B'C' in AA' and Oa is the reflection
of the circumcenter of A'B'C' in AA'.

Similarly Ob, Oc.

Oa, Ob, Oc are the reflections of the circumcenter of the cevian triangle of I in the bisectors AA',BB',CC', resp. Therefore the lines AOa, BOb, COc concur at the isogonal conjugate of the circumcenter of A'B'C'.

Coordinates ?

Generalization:
For P instead of I:

We have two problems:

1. The locus of P such that ABC, OaObOc are perspective. (Oa,Ob,Oc as defined above)
2. The locus of P such that ABC, QaQbQc are perspective, where Qa is the circumcenter of the triangle which is the reflection of A'B'C' in AA' (ie Qa is the reflection of the circumcenter of A'B'C' in AA') and similarly Qb,Qc

APH, 24 February 2012

----------------------------------------------------

Coordinates:

(a (a^5 b + a^4 b^2 - 2 a^3 b^3 - 2 a^2 b^4 + a b^5 + b^6 + a^5 c -
3 a^3 b^2 c - 2 a^2 b^3 c + 2 a b^4 c + 2 b^5 c + a^4 c^2 -
3 a^3 b c^2 - 2 a^2 b^2 c^2 - 3 a b^3 c^2 - b^4 c^2 - 2 a^3 c^3 -
2 a^2 b c^3 - 3 a b^2 c^3 - 4 b^3 c^3 - 2 a^2 c^4 + 2 a b c^4 -
b^2 c^4 + a c^5 + 2 b c^5 + c^6) : ...:...)

1. The locus is a 18 degree curve.

2. The locus is a 24 degree curve.

Nikos Dergiades, Hyacinthos #20870