Σάββατο 30 Μαρτίου 2013

ORTHOPOLAR CIRCLES

Let ABCD be a quadrilateral [quadragon], A',B',C',D' the circumcenters of BCD, CDA, DAB, ABC, resp. and P a point.

Denote:

1 = the orthopole of PA' wrt BCD

2 = the orthopole of PB' wrt CDA

3 = the orthopole of PC' wrt DAB

4 = the orthopole of PD' wrt ABC

Conjecture:

The points 1,2,3,4 are concyclic. The circle passes through the Poncelet point of ABCD (=the point where the NPCs of BCD, CDA, DAB, ABC concur)

The circle (1,2,3,4) is a line when ABCD is cyclic (ie A' = B' = C'= D' = O ==> PO = PA' = PB' = PC' := L, a line psiing through the circumcenter of the cyclic ABCD)

Antreas P. Hatzipolakis, 30 March 2013

Δεν υπάρχουν σχόλια:

Δημοσίευση σχολίου

ETC POINTS

X(73202) = DAO'S POINT OF ROUSSEL'S HEXAGON Barycentrics    Sin[A]*(2*Cos[A/3]*Cos[(2*(B - C))/3]*Cos[2*A - Pi/6] + Cos[(4*(B - C...