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Re: Antreas' Feuerbach concurrence
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Dear Antreas,
[APH], (with P replacing O)
>Let P be a point..
If P is on the Keipert hyperbola, then, the Brocard axes of the
>Let Ab and Ac be the orthogonal projections of the
>point A on the lines BP and CP. Similarly, define
>the points Bc and Ba, and the points Ca and Cb.
triangles AAbAc, BBcBa and CCaCb concur at the the center of the
hyperbola, X(115).
As well as the Keipert hyperbola, it looks like other loci are the
line at infinity, the sidelines and a degree 8.
The Euler lines of the triangles AAbAc, BBcBa and CCaCb concur for P
on the line at infinity and degrees 6 & 7.
Some ETC points,
1 > 11
4 > 125
13 > 115
14 > 115
74 > 125
80 > 11
Best regards
Peter. 0 Attachment
Dear Peter
[APH]:> >Let P be a point..
[PM]:
> >Let Ab and Ac be the orthogonal projections of the
> >point A on the lines BP and CP. Similarly, define
> >the points Bc and Ba, and the points Ca and Cb.
> If P is on the Keipert hyperbola, then, the Brocard axes of the
Very nice ! Thanks.
> triangles AAbAc, BBcBa and CCaCb concur at the the center of the
> hyperbola, X(115).
> As well as the Keipert hyperbola, it looks like other loci are the
> line at infinity, the sidelines and a degree 8.
>
> The Euler lines of the triangles AAbAc, BBcBa and CCaCb concur for P
> on the line at infinity and degrees 6 & 7.
> Some ETC points,
>
> 1 > 11 Feuerbach Hyp. center
> 4 > 125 Jerabek
> 13 > 115 Kiepert
> 14 > 115 Kiepert
> 74 > 125 Jerabek
> 80 > 11 Feuerbach
From these first points in ETC, we may conclude that the entire
locus of P [a 1+6+7 deg. curve] is transformed in a three point set:
Centers of Feuerbach,Jerabek, and Kiepert Hyperbolae.
But most likely the locus of the point of concurrence is the
entire NPC (+ some other things ?)
Greetings
Antreas 0 Attachment
In Hyacinthos message #10593, Peter Moses wrote:
>> The Euler lines of the triangles AAbAc, BBcBa
Is it possible that antigonal conjugates are mapped
>> and CCaCb concur for P on the line at infinity
>> and degrees 6 & 7.
>> Some ETC points,
>>
>> 1 > 11 Feuerbach Hyp. center
>> 4 > 125 Jerabek
>> 13 > 115 Kiepert
>> 14 > 115 Kiepert
>> 74 > 125 Jerabek
>> 80 > 11 Feuerbach
to the same point? In other words, if a point P has
the property that the Euler lines of triangles
AAbAc, BBcBa, CCaCb concur at one point Q, then the
antigonal conjugate P' of P also has this property,
and the Euler lines for P' concur at the same point
Q ? This looks pretty plausible to me from the
above data. (See Bernard Gibert's nice page
http://perso.wanadoo.fr/bernard.gibert/gloss/pointsandmapping.html
for a definition of antigonal conjugates.)
In Hyacinthos message #10594, Antreas P. Hatzipolakis
wrote:
>> From these first points in ETC, we may conclude
Hardly. Anyway, extraversion shows that since the
>> that the entire locus of P [a 1+6+7 deg. curve]
>> is transformed in a three point set:
>> Centers of Feuerbach,Jerabek, and Kiepert
>> Hyperbolae.
Feuerbach point X(11) serves as point of
concurrence, the three external Feuerbach points
must serve as points of concurrence, too. So we
would have, at least, 6 different points. But I
really suspect that the locus of the point of
concurrence is the whole ninepoint circle. If
anybody could prove or disprove that...
Darij Grinberg 0 Attachment
Dear Darij
[PM]:>>> The Euler lines of the triangles AAbAc, BBcBa
[DG]:
>>> and CCaCb concur for P on the line at infinity
>>> and degrees 6 & 7.
>>> Some ETC points,
>>>
>>> 1 > 11 Feuerbach Hyp. center
>>> 4 > 125 Jerabek
>>> 13 > 115 Kiepert
>>> 14 > 115 Kiepert
>>> 74 > 125 Jerabek
>>> 80 > 11 Feuerbach
>Is it possible that antigonal conjugates are mapped
Also, is it possible that the sextic and/or the septic
>to the same point? In other words, if a point P has
>the property that the Euler lines of triangles
>AAbAc, BBcBa, CCaCb concur at one point Q, then the
>antigonal conjugate P' of P also has this property,
>and the Euler lines for P' concur at the same point
>Q ? This looks pretty plausible to me from the
>above data.
of the locus is a selfantigonal curve?
That is, if P lies on the curve, then the antigonal of
P lies on the curve too ?
[APH]:>> From these first points in ETC, we may conclude
[DG]:
>> that the entire locus of P [a 1+6+7 deg. curve]
>> is transformed in a three point set:
>> Centers of Feuerbach,Jerabek, and Kiepert
>> Hyperbolae.
>Hardly. Anyway, extraversion shows that since the
Neither I believe that, since I wrote
(but you misquoted me!):
[APH]:>>But most likely the locus of the point of concurrence is the
Antreas
>>entire NPC (+ some other things ?)
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>[DG]:
[APH]:
>>Is it possible that antigonal conjugates are mapped
>>to the same point? In other words, if a point P has
>>the property that the Euler lines of triangles
>>AAbAc, BBcBa, CCaCb concur at one point Q, then the
>>antigonal conjugate P' of P also has this property,
>>and the Euler lines for P' concur at the same point
>>Q ? This looks pretty plausible to me from the
>>above data.
>Also, is it possible that the sextic and/or the septic
This is already contained in Darij's message, but I had not
>of the locus is a selfantigonal curve?
>That is, if P lies on the curve, then the antigonal of
>P lies on the curve too ?
read it carefully!
APH
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Dear Antreas and Peter,
> [APH]:
... which is bicircular and contains the points at infinity of the
> > >Let P be a point..
> > >Let Ab and Ac be the orthogonal projections of the
> > >point A on the lines BP and CP. Similarly, define
> > >the points Bc and Ba, and the points Ca and Cb.
>
> [PM]:
> > If P is on the Keipert hyperbola, then, the Brocard axes of the
> > triangles AAbAc, BBcBa and CCaCb concur at the the center of the
> > hyperbola, X(115).
> > As well as the Kiepert hyperbola, it looks like other loci are the
> > line at infinity, the sidelines and a degree 8.
Steiner ellipse (double points) and (I think) only five singular real
points A, B, C, X13, X14.
> >> The Euler lines of the triangles AAbAc, BBcBa and CCaCb concur for P
... the septic is Q030 and the sextic is tricircular containing the
> > on the line at infinity and degrees 6 & 7.
same five singular real points.
http://perso.wanadoo.fr/bernard.gibert/curves/q030.html
Q030 is indeed selfantigonal since its isogonal conjugate is itself an
inversible bicircular quintic which I will probably add to my web site.
Very nice indeed !
Best regards
Bernard
[Nontext portions of this message have been removed] 0 Attachment
Dear Antreas, Darij, Bernard, Peter and other Hyacinthists> > [APH]:
the
> > > >Let P be a point..
> > > >Let Ab and Ac be the orthogonal projections of the
> > > >point A on the lines BP and CP. Similarly, define
> > > >the points Bc and Ba, and the points Ca and Cb.
> >
> > [PM]:
> > > If P is on the Keipert hyperbola, then, the Brocard axes of the
> > > triangles AAbAc, BBcBa and CCaCb concur at the the center of
> > > hyperbola, X(115).
the
> > > As well as the Kiepert hyperbola, it looks like other loci are
> > > line at infinity, the sidelines and a degree 8.
real
>
> ... which is bicircular and contains the points at infinity of the
> Steiner ellipse (double points) and (I think) only five singular
> points A, B, C, X13, X14.
concur for P
>
> > >> The Euler lines of the triangles AAbAc, BBcBa and CCaCb
> > > on the line at infinity and degrees 6 & 7.
the
>
> ... the septic is Q030 and the sextic is tricircular containing
> same five singular real points.
itself an
>
> http://perso.wanadoo.fr/bernard.gibert/curves/q030.html
>
> Q030 is indeed selfantigonal since its isogonal conjugate is
> inversible bicircular quintic which I will probably add to my web
site.
The point is that the sextic has probably no other finite real point
than A,B,C and the Fermat points. Hence,
YOU CAN REMOVE THE SEXTIC FROM THE LOCUS.
Here is an explanation :
If PaPbPc is the pedal triangle of P, the direct similitude with
center Pc mapping the midpoint of AP to the midpoint of BP will map
the triangle AAbAc to BaBBc. It follows that the three triangles
AAbAc, BBcBa, CCaCb are equilateral at the same time, and this can
occur only when P is one of the Fermat points. I presume that your
sextic is the condition for one of these triangles to be
equilateral  something like OaGa = 0 where Oa and Ga are the
circumcenter and the centroid of AAbAc ; but, in this case, the
three Euler lines don't exist and, of course, cannot concur.
Now, an easy computation  starting, for instance, of the equation
xy=k for the rectangular hyperbola ABCP  gives the fact that, when
the three Euler lines concur, their common point is the center of
the hyperbola.
From this it follows that the locus of P for which the three Euler
lines concur is the union of the line at infinity (in which case,
the three lines are the same ones = GP) and of the curve of degree 7
above, in which case their common point, being the center of the
rectangular circumhyperbola going through P, lies on the NPCircle of
ABC.
Of course, we can now understand why the septic is selfantigonal
and why her isogonal conjugate  Bernard's quintic  is anallagmatic
wrt the circumcircle.
Friendly. JeanPierre 0 Attachment
>[APH]
We have seen the locus of P such that the Euler Lines of
>
>>Let P be a point..
>>Let Ab and Ac be the orthogonal projections of the
>>point A on the lines BP and CP. Similarly, define
>>the points Bc and Ba, and the points Ca and Cb.
the triangles AAbAc, BBcBa and CCaCb are concurrent.
Next natural question is:
Which is the locus of P such that the Euler Lines (variations: OK lines, or...)
of the triangles ABcCb, BCaAc and CAbBa are concurrent.
It seems that this is more complicated than the first locus!
Antreas
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>>[APH]
And since it is too complicated, let's see a special case:
>>
>>>Let P be a point..
>>>Let Ab and Ac be the orthogonal projections of the
>>>point A on the lines BP and CP. Similarly, define
>>>the points Bc and Ba, and the points Ca and Cb.
>
>We have seen the locus of P such that the Euler Lines of
>the triangles AAbAc, BBcBa and CCaCb are concurrent.
>
>Next natural question is:
>
>Which is the locus of P such that the Euler Lines (variations: OK
>lines, or...)
>of the triangles ABcCb, BCaAc and CAbBa are concurrent.
>
>It seems that this is more complicated than the first locus!
Let ABC be a triangle, L a line, and La, Lb, Lc
the parallels through A,B,C to the line L, resp.
Bc := the orthogonal projection of B on Lc
Cb := the orthogonal projection of C on Lb
Similarly Ca,Ac; Ab, Ba
The Euler lines of ABcCb, BCaAc, CAbBa [and ABC ?] are concurrent (?)
Antreas
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Dear Antreas,
[APH]>>Let P be a point..
Yes, I think so.
>>Let Ab and Ac be the orthogonal projections of the
>>point A on the lines BP and CP. Similarly, define
>>the points Bc and Ba, and the points Ca and Cb.
> Next natural question is:
>
> Which is the locus of P such that the Euler Lines (variations: OK
> lines, or...) of the triangles ABcCb, BCaAc and CAbBa are
> concurrent.
>
> It seems that this is more complicated than the first locus!
I reckon that the Euler line locus is L.inf + degree 17 and Brocard
a degree 18.
I suspect X(4) lives on both locii.
Best regards
Peter. 0 Attachment
Dear Peter
>[APH]
[PM]:
>>>Let P be a point..
>>>Let Ab and Ac be the orthogonal projections of the
>>>point A on the lines BP and CP. Similarly, define
>>>the points Bc and Ba, and the points Ca and Cb.
>
>> Next natural question is:
>>
>> Which is the locus of P such that the Euler Lines (variations: OK
>> lines, or...) of the triangles ABcCb, BCaAc and CAbBa are
>> concurrent.
>>
>> It seems that this is more complicated than the first locus!
>Yes, I think so.
Yes, X(4) = H, lives on both:
>I reckon that the Euler line locus is L.inf + degree 17 and Brocard
>a degree 18.
>I suspect X(4) lives on both locii.
If AA', BB', CC' are the cevians of H, then Ab = Cb = B', etc
and we know that the Euler Lines / Brocard axes of
AB'C', BC'A', CA'B' are concurrent.
Greetings
Antreas
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[APH]:>Let ABC be a triangle, L a line, and La, Lb, Lc
The lines BcCb, CaAc, AbBa concur at a point on the NPC of ABC.
>the parallels through A,B,C to the line L, resp.
>
>Bc := the orthogonal projection of B on Lc
>Cb := the orthogonal projection of C on Lb
>
>Similarly Ca,Ac; Ab, Ba
>The Euler lines of ABcCb, BCaAc, CAbBa [and ABC ?] are concurrent (?)
It is an old result I read in a Greek book and have posted
in Hyacinthos (search the archive with keyword Panakis)
If the line L is the Euler line, or the Brocard axis or the OI
line of ABC, then which is the point of concurrence of the lines
BcCb, CaAc, AbBa ?
APH
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>[APH]
Which is the locus of P such that ABC,
>
>>Let P be a point..
>>Let Ab and Ac be the orthogonal projections of the
>>point A on the lines BP and CP. Similarly, define
>>the points Bc and Ba, and the points Ca and Cb.
1. Triangle bounded by (AbAc, BcBa, CaCb)
2. Triangle bounded by (BcCb, CaAc, AbBa)
are perspective?
[Special Cases: (AbAc, BcBa, CaCb), (BcCb, CaAc, AbBa)
be concurrent]
APH
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Dear Antreas,
> >[APH]
1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
> >
> >>Let P be a point..
> >>Let Ab and Ac be the orthogonal projections of the
> >>point A on the lines BP and CP. Similarly, define
> >>the points Bc and Ba, and the points Ca and Cb.
>
> Which is the locus of P such that ABC,
>
> 1. Triangle bounded by (AbAc, BcBa, CaCb)
>
> 2. Triangle bounded by (BcCb, CaAc, AbBa)
>
> are perspective?
X(1) > X(2)
X(3) > 1/((a^2  SA) SA + 3 SB SC) : :
X(4) > X(4)
P on circumcircle > point on NPC midway between P and X(4)
2/. DARBOUX CUBIC + L.inf + circumcircle + sides.
X(1) > X(7)
X(3) > X(184)
X(4) > X(4)
X(84) > X(1857)
Best regards,
Peter. 0 Attachment
Dear Antreas and Peter> > >>Let P be a point..
I presume that your ellipses are the circles with diameter BC, CA
> > >>Let Ab and Ac be the orthogonal projections of the
> > >>point A on the lines BP and CP. Similarly, define
> > >>the points Bc and Ba, and the points Ca and Cb.
> >
> > Which is the locus of P such that ABC,
> >
> > 1. Triangle bounded by (AbAc, BcBa, CaCb)
> >
> > 2. Triangle bounded by (BcCb, CaAc, AbBa)
> >
> > are perspective?
>
> 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
> X(1) > X(2)
> X(3) > 1/((a^2  SA) SA + 3 SB SC) : :
> X(4) > X(4)
> P on circumcircle > point on NPC midway between P and X(4)
and AB, corresponding to the case where two lines are the same ones.
Friendly. JeanPierre 0 Attachment
Dear JeanPierre & Antreas,
[APH]>>Let P be a point..
[JPE]
>>Let Ab and Ac be the orthogonal projections of the
>>point A on the lines BP and CP. Similarly, define
>>the points Bc and Ba, and the points Ca and Cb.
>>
>> Which is the locus of P such that ABC,
>>
>> 1. Triangle bounded by (AbAc, BcBa, CaCb)
>>
>> 2. Triangle bounded by (BcCb, CaAc, AbBa)
>>
>> are perspective?
>[PM]
> 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
> X(1) > X(2)
> X(3) > 1/((a^2  SA) SA + 3 SB SC) : :
> X(4) > X(4)
> P on circumcircle > point on NPC midway between P and X(4)
>
> I presume that your ellipses are the circles with diameter BC, CA
Yes!, I think so. I should have stared at my diagram a bit longer:)
> and AB, corresponding to the case where two lines are the same
> ones.
Best regards,
Peter. 0 Attachment
Dear Peter and JeanPierre
[PM]:>> 1/. DARBOUX CUBIC + L.inf + circumcircle + 3 ellipses.
[JPE]
>> I presume that your ellipses are the circles with diameter BC, CA
[PM]:
>> and AB, corresponding to the case where two lines are the same
>> ones.
>Yes!, I think so. I should have stared at my diagram a bit longer:)
In other loci, we have seen these circles with the Neuberg cubic + Linf + (O)
How about a locus problem where these circles come with an other isogonal
cubic?
APH
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Dear Antreas,
>>[APH]
Call the triangle bounded by (AbAc, BcBa, CaCb), A1B1C1
>>
>>Let P be a point..
>>Let Ab and Ac be the orthogonal projections of the
>>point A on the lines BP and CP. Similarly, define
>>the points Bc and Ba, and the points Ca and Cb.
Call the triangle bounded by (BcCb, CaAc, AbBa), A2B2C2
Circumcircles of A1B1C1 and A2B2C2 meet the NPC of ABC at Q,
x (a^2 y  b^2 y + c^2 y  a^2 z  b^2 z + c^2 z)
(c^2 x y  b^2 x z  b^2 y z + c^2 y z) : :
the center of the hyperbola ABCHP.
Circumcircle of A2B2C2 & NPC of ABC are tangent.
A3 = BBc /\ CCb = H of BCP
B3 = CCa /\ AAc = H of CAP
C3 = AAb /\ BBa = H of ABP
A3,B3 & C3 are on the above hyperbola, opening the familiar flood
gate of NP circles meeting at Q.
NPC of A3BC passes through A1, Bc and Cb .. and Q
NPC of AB3C3 passes through A1, AAc/\BcBa, AAb/\CaCb .. and Q
NPC A3BC /\ NPC A3BC3 at midpoint BA3 on BBc .. and Q.
Best regards,
Peter. 0 Attachment
[APH #10485]:> Let ABC be a triangle and A'B'C' the cevian
Let MaMbMc be the medial triangle of ABC.
> triangle of I.
>
> Ab := the Orthogonal projection of A on BB'
> Ac := the Orthogonal projection of A on CC'
>
> Similarly Bc,Ba; Ca,Cb
>
> The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
> the Feuerbach point (??)
AbAc is the sideline MbMc of the medial triangle.
So the problem in other words is:
Let ABC be a triangle, A'B'C' the cevian
triangle of I and MaMbMc the cevian triangle of G.
The perpendicular from A to BB' intersects
the line MbMc at Ab.
The perpendicular from A to CC' intersects
the line MbMc at Ac.
Similarly Bc,Ba; Ca,Cb
The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
the Feuerbach point.
I was wondering what happens if we draw perpendiculars to
the cevians of I, not from A,B,C, but from Ma,Mb,Mc.
That is:
Let ABC be a triangle, A'B'C' the cevian
triangle of I and MaMbMc the cevian triangle of G.
The perpendicular from Ma to BB' intersects
the line MbMc at Ab.
The perpendicular from Ma to CC' intersects
the line MbMc at Ac.
Similarly Bc,Ba; Ca,Cb
The Euler Lines of MaAbAc, MbBcBa, McCaCb are concurrent
(at some point on IG ???)
Antreas 0 Attachment
[APH]> Let ABC be a triangle, A'B'C' the cevian
In general: For MaMbMc = a triangle homothetic to
> triangle of I and MaMbMc the cevian triangle of G.
>
> The perpendicular from Ma to BB' intersects
> the line MbMc at Ab.
>
> The perpendicular from Ma to CC' intersects
> the line MbMc at Ac.
>
> Similarly Bc,Ba; Ca,Cb
>
> The Euler Lines of MaAbAc, MbBcBa, McCaCb are concurrent
> (at some point on IG ???)
ABC with homothety (G,t).
APH 0 Attachment
Dear Antreas,
> Let ABC be a triangle, A'B'C' the cevian
Yes, you are right! In fact, we can say more: The three triangles
> triangle of I and MaMbMc the cevian triangle of G.
>
> The perpendicular from Ma to BB' intersects
> the line MbMc at Ab.
>
> The perpendicular from Ma to CC' intersects
> the line MbMc at Ac.
>
> Similarly Bc,Ba; Ca,Cb
>
> The Euler Lines of MaAbAc, MbBcBa, McCaCb are concurrent
> (at some point on IG ???)
share the same circumcenter. It's the Spieker point Sp of ABC.

Here's a proof:
Note that MaAbAc is taken to a familiar triangle by a homothecy g
with factor 1 and center the midpoint of MbMc. It's the triangle of
the initial problem, which has his circumcenter at the midpoint of
AI. Now consider the homothecy h with center A and factor 2, so we
know that the circumcenter of MaAbAc is mapped to I by gh.
On the other hand, gh is a homothecy with center G and factor 2, so
it maps the incenter of MaMbMc to the incenter of ABC, which is I.
This shows that Sp, the incenter of MaMbMc coincides with the
circumcenter of MaAbAc.

Unfortunately, I think this proof is worthless for your general
theorem, which seems to be true! Amazing...
All the best,
Jan 0 Attachment
[APH]> Let ABC be a triangle and A'B'C' the cevian
Let La, Lb, Lc be the Reflections of the Euler Lines of
> triangle of I.
>
> Ab := the Orthogonal projection of A on BB'
> Ac := the Orthogonal projection of A on CC'
>
> Similarly Bc,Ba; Ca,Cb
>
> The Euler Lines of AAbAc, BBcBa, CCaCb are concurrent at F,
> the Feuerbach point (??)
AAbAc, BBcBa, CCaCb in the angle bisectors of A, B, C, resp.
and Ma, Mb, Mc the reflections of La, Lb, Lc in the
corresponding angle bisectors of the Excentral triangle
of ABC.
I think that:
1. La, Lb, Lc
2. Ma, Mb, Mc
are concurrent.
Points?
Antreas
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