Browse Groups

• 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
View Source
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 1 of 3 , Sep 3, 2001
View Source
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 1 of 3 , Sep 3, 2001
View Source
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
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.