[Antreas P. Hatzipolakis]:
[Peter Moses]:
Generalization of Romantics of Geometry Problem 816
Let ABC be a triangle, P, Q two isogonal conjugate points and ApBpCp, AqBqCq the pedal triangles of P, Q, resp.
Denote:
A' = BpCq Intersection BqCp.
A' lies on the line PQ.
Problem:
Similarly B' = CpAq Inters. CqAp and C' = ApBq Inters. AqBp
Which point is the centroid of the degenerated triangle A'B'C' in terms of the coordinates of P?
Let ABC be a triangle, P, Q two isogonal conjugate points and ApBpCp, AqBqCq the pedal triangles of P, Q, resp.
Denote:
A' = BpCq Intersection BqCp.
A' lies on the line PQ.
Problem:
Similarly B' = CpAq Inters. CqAp and C' = ApBq Inters. AqBp
Which point is the centroid of the degenerated triangle A'B'C' in terms of the coordinates of P?
[Peter Moses]:
Hi Antreas,
They seem rather long. However, the point will lie on the line P,Q. So for O and H, it will be on Euler at:
a^10 b^2-a^8 b^4-4 a^4 b^8+7 a^2 b^10-3 b^12+a^10 c^2-4 a^8 b^2 c^2+3 a^6 b^4 c^2+6 a^4 b^6 c^2-16 a^2 b^8 c^2+10 b^10 c^2-a^8 c^4+3 a^6 b^2 c^4-6 a^4 b^4 c^4+9 a^2 b^6 c^4-13 b^8 c^4+6 a^4 b^2 c^6+9 a^2 b^4 c^6+12 b^6 c^6-4 a^4 c^8-16 a^2 b^2 c^8-13 b^4 c^8+7 a^2 c^10+10 b^2 c^10-3 c^12::
on lines {2,3}
For P = X(30) or X(74) we end up with X(9140).
Best regards,
Peter Moses.
Δεν υπάρχουν σχόλια:
Δημοσίευση σχολίου