## Re: [EMHL] Re: Pediac triangles

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• Dear Floor and Jean-Pierre, ... Aren t these triangles what I called P-ac triangles (as generalizations of H-ac = orthiac triangles)? ... (Re: [EMHL]
Message 1 of 3 , Sep 3, 2001
Dear Floor and Jean-Pierre,

[FvL]:
>> Let P be a point, A'B'C' its pedal triangles. From A' drop
>> perpendicular
>> altitudes to the sides AC and AB, ending in Ab and Ac respectively.
>> Then we could call A'AbAc pediac and AAbAc synpediac triangles.

Aren't these triangles what I called P-ac triangles (as generalizations
of H-ac = orthiac triangles)?

| Let's now replace H with another point P:
| Let ABC be a triangle and A'B'C' the pedal triangle of P.
| We define the P-ac triangles A'AbAc, B'BcBa, C'CaCb
| and Syn-P-ac triangles AAbAc, BBcBa, CCaCb.
(Re: [EMHL] Synorthiac triangles)

If they are the same, then I prefer Floor's term Pediac than the
abbreviated mine (P-ac).

[JPE]:
>Here is a little problem related to those triangles :
>Construct P such as the three circumcircles of the pediac triangles
>of P (or synpediac triangles, they are the same ones) have the same
>In other word, construct P such as AA' = BB' = CC'.

I remember this variation:

Draw three CONCURRENT cevians AA', BB', CC' such that
AA' = BB' = CC'.

I will look at F.G.-M., and give the precise reference at
the end of this message.

[JPE]:
>I suppose that the solution this problem is well known, but not by me.
>I'm pretty sure that there are exactly two solutions P, P', that H is
>the midpoint of PP' and that the line PP' is parallel to the major
>axis of a Steiner ellipse (thus this line is the Steiner line of a
>common point of the circumcircle and the line OK).
>Starting from that, it should not be very difficult to finish.

I found the reference:

On peut se proposer de determiner trois ceviennes concourantes qui
aient meme longueur totale.

Si F est leur point de concours, ce point est le foyer de la conique
circonscrite au triangle donne, et qui a pour centre de centre de
gravite du triangle (L. BICKART, I. M., 1910, p. 170, no 3724) Voir comme
reponse, p. 256 a 260, art. par MM. WELSCH, E. MALO, etc.

F.G.-M.: Exercices de geometrie. 5e ed. (1912), paragraph 1242. l, #4

Antreas
• Dear Antreas, ... [APH] ... [APH] ... Yes, they are the same, but I didn t seem to notice that. Let s call them Pediac and SynPediac from now. Kind regards,
Message 2 of 3 , Sep 3, 2001
Dear Antreas,

> [FvL]:
> >> Let P be a point, A'B'C' its pedal triangles. From A' drop
> >> perpendicular
> >> altitudes to the sides AC and AB, ending in Ab and Ac respectively.
> >> Then we could call A'AbAc pediac and AAbAc synpediac triangles.

[APH]
> Aren't these triangles what I called P-ac triangles (as generalizations
> of H-ac = orthiac triangles)?

[Quote of APH]:
> | Let's now replace H with another point P:
> | Let ABC be a triangle and A'B'C' the pedal triangle of P.
> | We define the P-ac triangles A'AbAc, B'BcBa, C'CaCb
> | and Syn-P-ac triangles AAbAc, BBcBa, CCaCb.
> (Re: [EMHL] Synorthiac triangles)

[APH]
> If they are the same, then I prefer Floor's term Pediac than the
> abbreviated mine (P-ac).

Yes, they are the same, but I didn't seem to notice that. Let's call
them Pediac and SynPediac from now.

Kind regards,
Sincerely,
Floor.
• Dear Antreas and other Hyacinthists, ... (where A B C is the pedal triangle of P) ... me. ... is ... [APH] ... comme ... #4 Translation : There are two points
Message 3 of 3 , Sep 3, 2001
Dear Antreas and other Hyacinthists,

> [JPE]:
> >Here is a little problem related to those triangles :
> >Construct P such as the three circumcircles of the pediac triangles
> >of P (or synpediac triangles, they are the same ones) have the same
> >In other word, construct P such as AA' = BB' = CC'.
(where A'B'C' is the pedal triangle of P)
> >I suppose that the solution this problem is well known, but not by
me.
> >I'm pretty sure that there are exactly two solutions P, P', that H
is
> >the midpoint of PP' and that the line PP' is parallel to the major
> >axis of a Steiner ellipse (thus this line is the Steiner line of a
> >common point of the circumcircle and the line OK).
> >Starting from that, it should not be very difficult to finish.

[APH]
> I found the reference:
>
> On peut se proposer de determiner trois ceviennes concourantes qui
> aient meme longueur totale.
>
> Si F est leur point de concours, ce point est le foyer de la conique
> circonscrite au triangle donne, et qui a pour centre de centre de
> gravite du triangle (L. BICKART, I. M., 1910, p. 170, no 3724) Voir
comme
> reponse, p. 256 a 260, art. par MM. WELSCH, E. MALO, etc.
>
> F.G.-M.: Exercices de geometrie. 5e ed. (1912), paragraph 1242. l,
#4

Translation : There are two points with equal cevians : the focii F,
F' of the Steiner circumellipse.
Many thanks, Antreas for the reference.

So, there are two points P such as AA' = BB' = CC' where A'B'C' is
the pedal triangle of P; those points are homothetic of F, F' in the
homothecy (L, 3/2) - L = de Longchamps point - and, of course, we can
now construct those points.

Friendly. Jean-Pierre
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