- Apr 23, 2014We have again concyclic points?The six points are concyclic. The circumcenters of the circles, as P's move on the Euler lines, being same,(ie have same coordinates with respect to six triangles ie they are all circumcenters, orthocenters, NPC centers etc)2Pa,2Pb, 2Pc three points on the Euler lines eX, eY, eZ of the case 2, such that all six points are sameIn general:The six circumcenters are concyclic. The same is true for the NPC centers of the triangles.

Let X,Y,Z be three lines.Denote:xL1, xL2, xL3 = the parallels to X through A,B,C, resp.

yL1, yL2, yL3 = the parallels to Y through A,B,C, resp.

zL1, zL2, zL3 = the parallels to Z through A,B,C, resp.

xAxBxC = the triangle bounded by the reflections of xL1, xL2, xL3 in BC,CA,AB, resp.

yAyByC = the triangle bounded by the reflections of yL1, yL2, yL3 in BC,CA,AB, resp.

zAzBzC = the triangle bounded by the reflections of zL1, zL2, zL3 in BC,CA,AB, resp.

eX, eY, eZ = the Euler lines of xAxBxC, yAyByC, zAzBzC, resp.

rX, rY, rZ = the radical axis of (O) and (N) of the triangles xAxBxC, yAyByC, zAzBzC, resp.

We have seen that xE, yE, zE pass through a fixed point, and also xR, yR, zR pass through an other fixed point.

Now,

xeX = the parallel through A to the line Euler line eX

yeY = the parallel through B to the line eY

zeZ = the parallel through C to the line eZ

xrX = the parallel through A to the line (radical axis) rX

yrY = the parallel through B to the line rY

zrZ = the parallel through C to the line rZ

Question: For which lines X,Y,Z the lines xeX, yeY, zeZ and also the lines

xrX, yrY, zrZ are concurrent?

Some examples of loci:

Two of the lines X,Y,Z are fixed and the third is variable (trilinear polar of a variable

point P). Or we can take X,Y,Z = cevians of a variable point P or sidelinesof a variable triangle (pedal or cevian triangle of a point P).

Two special cases of lines X.Y,Z:

The lines X,Y,Z are:

1. the sidelines BC,CA,AB, resp.

2. the altitude lines a_altitude, b_altitude, c_altitude, resp..

[in the first case we have X = BC = xL2 = xL3 etc and in the second X = xL1 etc]Observations:Let 1Oa, 1Ob,1Oc be the circumcenters of xAxBxC,yAyByC, zAzBzC, resp. for case 1, and 2Oa,2Ob,2Oc for case 2.

For case 1. The lines xeX, yeY, zeZ concur on a point Q1 on the circumcircle of ABC

and the lines xrX, yrY, zrZ on another point Q2 on the circumcircle.

For case 2. The lines xeX, yeY, zeZ concur on a point T1 on the circumcircle of ABC

and the lines xrX, yrY, zrZ on an other point T2 on the circumcircle.

We have Q1 = T2 and Q2 = T1

Let 1Pa, 1Pb, 1Pc be three points on the Euler lines eX, eY, eZ of the case 1 and

are collinear on a line k, and have same positions on the line k as the P's on the Euler lines.[this means, for exampe, that if O',N',H' are the circumcenters on the line k corresponding to Os, Ns, H's on the Euler lines,then |N'O'| = |N'H'| as |NO| = |NH|)Note: How about if the lines of the cases 1 and 2 are perpendiculars in general but not necessarily sidelines and altitudes?

Hope there are no too many errors, typos etc .... :-)Antreas - Next post in topic >>