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