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

Another relationship between Napoleon cubic and Neuberg cubic

Another relationship between Napoleon cubic K005 and Neuberg cubic K001 The world of Triangle Geometry is very intrincate. There are many...