HYACINTHOS 22717

Antreas P. Hatzipolakis

 

 

I was wondering what we get if we draw perpendiculars to trisectors

of a triangle like the perpendiculars to cevians in the antipedal triangle
configuration.

In details:

Let ABC be a triangle:

Denote:

AAb,AAc = the trisectors of A with the traces Ab, Ac near to B, C, resp.

BBc,BBa = the trisectors of B with the traces Bc, Ba near to C, A, resp.

CCa,CCb = the trisectors of C with the traces Ca, Cb near to A, B, resp.

ab, ac = the perpendicular lines to trisectors AAb, AAc at A, resp

bc, ba = the perpendicular lines to trisectors BBc, BBa at B, resp.

ca, cb = the perpendicular lines to trisectors CCa, CCb at C, resp.

The six perpendiculars form three quadrilaterals respective to A, B, C.
namely:

Q1 = : (bc, ba, ca, cb) respective to A

Q2 = : (ca, cb, ab, ac) respective to B

Q3 = : (ab, ac, bc, ba) respective to C

Two general questions:

Let P be a quadrilateral point and Pa, Pb, Pc the same quadrilateral

points of the quadrilaterals Q1,Q2, Q3, resp..

Let L be a quadrilateral line and La, Lb, Lc the same quadrilateral
lines of the quadrilaterals Q1,Q2,Q3,resp.

For which P's, llisted in EQF (*), the triangles ABC, PaPbPc are perspective?

For which L's, llisted in EQF (*),  the lines La,Lb,Lc are concurrent

(or more generally, bound a triangle perspective with ABC)?

The simlest points P are the Miquel point and the center of the Miquel circle.

and the simplest line is the Newton line.

Is the triangle ABC perspective with the triangles with vertices these

simple points of the quadrilaterals Q1,Q2,Q3 and are concurrent their
Newton lines ?

(*)
http://chrisvantienhoven.nl/index.php/mathematics/encyclopedia

APH

 

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