Δευτέρα, 1 Σεπτεμβρίου 2014

RADICAL CENTER - EULER LINE

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

Denote:

Oab, Oac = the circumcenters of AA'B, AA'C, resp.

(O1) = the circumcircle of A'OabOac. Similarly (O2), (O3).

P* = the radical center of (O1),(O2), (O3)

For P = G, the G* is lyimg on the Euler line of ABC.

Locus: Which is the locus of P such that P* is lying 1. on the OP line 2. on the Euler Line of ABC?

Antreas P. Hatzipolakis, 1 September 2014

Peter Moses:

G* = X(140) = midpoint of ON

GENERALIZATION

Let P is a point on Euler line of triangle ABC and DEF is pedal triangle of P. Let Oab,Oac be the circumcenters of triangle DAB,DAC and (Oa) is circumcircle of triangle DOabOac. Similarly, we have circle (Ob),(Oc). Then radical center of (Oa),(Ob),(Oc) lies on Euler line, too.

Tran Quang Hung.

Response


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