## Re: [EMHL] GpP // GH

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• Dear Jean-Pierre, [APH] ... Hmmm... If one tries it analyticaly, he has to work a week, I am afraid! Let P = (x:y:z) in normals. 1st step: Computation of the
Message 1 of 1 , Aug 31 4:19 PM
Dear Jean-Pierre,

[APH]
>> Let ABC be a triangle and Gp the centroid of the pedal triangle of P.
>> Which is the locus of P such that the line PGp is parallel to the
>> Euler line of ABC?

[JPE]:
>P -> Gp is affine with K (symedian) as fixed point. Thus, the locus
>of P such as PGp has a fixed direction is allways a line through K.
>For P = O, PGp is obviously the Euler line. Thus your locus is the
>Brocard axis OK.

Hmmm... If one tries it analyticaly, he has to work a week, I am afraid!

Let P = (x:y:z) in normals.
1st step: Computation of the normals of Gp.
They are (2x + ycosC + zcosB : xcosC + 2y + zcosA : xcosB + ycosA + 2z)

2nd step: Formulation of the equation of the line PGp. It is:
(y^2cosA - z^2cosA + xycosB - xzcosC)X +
(z^2cosB - x^2cosB + yzcosC - yxcosA)Y +
(x^2cosC - y^2cosC + zxcosA - zycosB)Z = 0

3rd step: Finding the condition to be the lines:
line at oo: xsinA + cyclic = 0, Euler line: xsin2Asin(B-C) + cyclic = 0,
and PGp concurrent.

They are concurrent iff

| sinA sinB sinC |
| |
| sin2Asin(B-C) sin2Bsin(C-A) sin2Csin(A-B) | = 0
| |
| y^2cosA - z^2cosA + xycosB - xzcosC [no space for the rest!]

.... and no time for more computations!

[I believe that if one makes them, then will get the line at oo and your OK]

Thanks !

Antreas
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