Πέμπτη, 14 Μαρτίου 2013

REFLENTING PERPENDICULARS

Let ABC be a triangle and A'B'C' the pedal triangle of H (orthic triangle).

Denote:

La = the reflection of HA' in B'C'

Lb = the reflection of HB' in C'A'

Lc = the reflection of HC' in A'B'

Ma = the reflection of La in OA.

Mb = the reflection of Lb in OB.

Mc = the reflection of Lb in OC.

The lines Ma,Mb,Mc concur at a point Q. Denote:

A" = BC /\ Ma

B" = CA /\ Mb

C" = AB /\ Mc

The triangles ABC, A"B"C" are orthologic, with orthologic centers Q,S.

Locus:

Let ABC be a triangle P,P* two isogonal conjugate points and A'B'C' the pedal triangle of P.

Denote:

La = the reflection of PA' in B'C'

Lb = the reflection of PB' in C'A'

Lc = the reflection of PC' in A'B'

Ma = the reflection of La in AP*.

Mb = the reflection of Lb in BP*.

Mc = the reflection of Lb in CP*.

Denote:

A" = BC /\ Ma

B" = CA /\ Mb

C" = AB /\ Mc

(Ma,Mb,Mc are perpendiculars to BC,CA,AB, resp.)

Which is the locus of P such that:

1. ABC, triangle bounded by (Ma,Mb,Mc) are perspective? Special case: Ma,Mb,Mc be concurrent.

2. ABC, A"B"C" are orthologic?

Antreas P. Hatzipolakis, 14 March 2013

2 σχόλια:

  1. Q=-4*b^4*a^6-2*c^2*a^6*b^2-4*c^4*a^6-2*a^10+2*b^6*c^4-3*c^2*b^8+c^10-3*c^8*b^2-2*c^8*a^2+2*c^6*a^4+5*a^8*c^2+b^10+2*c^6*b^4-2*c^4*b^2*a^4+2*c^6*b^2*a^2+2*c^2*b^6*a^2-2*c^2*b^4*a^4-2*b^8*a^2+2*b^6*a^4+5*a^8*b^2 : :
    (Barycentrics)

    Cesar Lozada

    ΑπάντησηΔιαγραφή
  2. More simple form (barycentrics)
    -2*a^10+(5*(c^2+b^2))*a^8-(c^2*(b^2+4*c^2)+b^2*(4*b^2+c^2))*a^6+(2*(c^2+b^2))*(b^2-c^2)^2*a^4+(b^2-c^2)^2*((c^2+b^2)^2-b^2*c^2)*a^2+(c^2+b^2)*(b^2-c^2)^4 : :

    ΑπάντησηΔιαγραφή

 

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