HYACINTHOS 22720

 

 

 

[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,

 

And here is the case with perpendiculars drawn through the vertices at the trisectors.

Let ABC be the reference triangle.

Let ab2 be the perpendicular at the trisector through A nearest to B.

Define ac2, bc2, ba2, ca2, cb2 accordingly.

Let Q1a be the Miquel Point (QL-P1) of the Quadrilateral formed by bc2, ba2, ca2, cb2.

Let Q1b be the Miquel Point (QL-P1) of the Quadrilateral formed by ca2, cb2, ab2, ac2.

Let Q1c be the Miquel Point (QL-P1) of the Quadrilateral formed by ab2, ac2, bc2, ba2.

Now Q1a, Q1b, Q1c  lie on the circumcircle of ABC.

And Q1a.Q1b.Q1c  is perspective with ABC with perspector N1.

 

The trilinear coordinates of N1 are:

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

Of course there are 3 sets of trisectors.

So there is also a 2^nd  and 3^rd point, resp. N2 and N3 with trilinear coordinates:

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

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

The search values of N1, N2, N3 are resp.

M1: 2.89615190425

M2: -0.91057320180

M3: 7.122469166

 

N1 lies on these lines:

X(3).X(358),

X(357).X(3278),

X(1136).X(3334)

X(3281),X(3602)

X(3335),X(3604)

 

N2 lies on these lines:

X(3).X(1135),

X(357), X(3334)

X(1134), X(3282)

X(3279),X(3604)

X(3335),X(3603)

 

N3 lies on these lines:

X(3).X(1137),

X(1136), X(3280)

X (1134), X (3334)

X(3283),X(3603)

X(3335),X(3602)

 

 

Best regards,

 

Chris van Tienhoven

www.chrisvantienhoven.nl

 

p.s.

I recalculated the lines of my former message with more ETC centers

M1 lies on these lines:

X(3).X(358),

X(357).X(3279),

X(1136),X(3335)

X(3334),X(3604)

 

M2 lies on these lines:

X(3).X(1135),

X(357).X(3335),

X(1134), X(3283)

X(3278),X(3604)

X(3334),X(3603)

 

M3 lies on these lines:

X(3).X(1137),

X(1134), X(3335)

X(1136), X(3281)

X(3282),X(3603)

X(3334),X(3602)