- Dear Floor and Jean-Pierre,

[FvL]:>> Let P be a point, A'B'C' its pedal triangles. From A' drop

Aren't these triangles what I called P-ac triangles (as generalizations

>> 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.

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 :

I remember this variation:

>Construct P such as the three circumcircles of the pediac triangles

>of P (or synpediac triangles, they are the same ones) have the same

>radius.

>In other word, construct P such as AA' = BB' = CC'.

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 found the reference:

>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.

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,

> [FvL]:

[APH]

> >> 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

[Quote of APH]:

> of H-ac = orthiac triangles)?

> | Let's now replace H with another point P:

[APH]

> | 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

Yes, they are the same, but I didn't seem to notice that. Let's call

> abbreviated mine (P-ac).

them Pediac and SynPediac from now.

Kind regards,

Sincerely,

Floor. - Dear Antreas and other Hyacinthists,

> [JPE]:

(where A'B'C' is the pedal triangle of P)

> >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

> >radius.

> >In other word, construct P such as AA' = BB' = CC'.

> >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

[APH]

> >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:

comme

>

> 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

> reponse, p. 256 a 260, art. par MM. WELSCH, E. MALO, etc.

#4

>

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

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