HYACINTHOS 22719

 

 

 

[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

 

[Chris van Tienhoven]:

Dear Antreas,

 

You really know where the treasures are hidden!

 

I simplify and amend your way of research this way:

Let ABC be the reference triangle.

Denote ab = the trisector through A nearest to B.

Define ac, bc, ba, ca, cb accordingly.

Let P1a be the Miquel Point (QL-P1) of the Quadrilateral formed by bc, ba, ca, cb.

Let P1b be the Miquel Point (QL-P1) of the Quadrilateral formed by ca, cb, ab, ac.

Let P1c be the Miquel Point (QL-P1) of the Quadrilateral formed by ab, ac, bc, ba.

Now P1a, P1b, P1c  lie on the circumcircle of ABC.

And P1a.P1b.P1c  is perspective with ABC with perspector M1.

 

The trilinear coordinates of M1 are:

Sin[A/3] Sin[(2 A)/3] : :

Of course there are 3 sets of trisectors.

So there is also a 2^nd  and 3^rd point, resp. M2 and M3 with trilinear coordinates:

Sin[(A + 2 Pi)/3] Sin[2(A + 2 Pi)/3]  : :

Sin[(A + 4 Pi)/3] Sin[2(A + 4 Pi)/3]  : :

The search values of M1, M2, M3 are resp.

M1: 0.19546078898

M2: 2.55522134223

M3: 6.6496360194

 

M1 lies on these lines:

X(3).X(358),

X(357).X(3279),

X(1136).X(3335).

 

M2 lies on these lines:

X(3).X(1135),

X(1134), X(3283)

X(357), X(3335)

 

M3 lies on these lines:

X(3).X(1137),

X(1136), X(3281)

X (1134), X (3335)

 

Best regards,

 

Chris van Tienhoven

www.chrisvantienhoven.nl

 

p.s.

1. I didn’t use your method of using perpendiculars to the trisectors because the method without perpendiculars even seemed better.
2. The method of using perpendiculars is not quite new. See remarks in ETC at X(5628).