## The Ehrmann points of a triangle

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• Let ABC be a triangle and AA , BB , CC its altitudes. Let Ab, Ac, Bc, Ba, Ca, Cb be the projections of A , A , B , B , C , C on AB, CA, BC, AB, CA, BC,
Message 1 of 14 , Aug 1, 2003
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Let ABC be a triangle and AA', BB', CC' its altitudes. Let Ab,
Ac, Bc, Ba, Ca, Cb be the projections of A', A', B', B', C',
C' on AB, CA, BC, AB, CA, BC, respectively.

It is well-known that the points Ab, Ac, Bc, Ba, Ca, Cb lie on
the Taylor circle of triangle ABC.

It is also important that the lines AbAc, BcBa, CaCb are
antiparallel to BC, CA, AB, whereas the lines BaCa, CbAb,
AcBc are parallel to BC, CA, AB.

Triangles AB'C' and AAcAb are homothetic; triangles HB'C' and
A'AcAb are homothetic, where H is the orthocenter of triangle
ABC.

We have the following theorems:

(1a) The Euler lines of triangles HB'C', HC'A', HA'B' concur
at one point, namely the Schiffler point of triangle
A'B'C'. It lies on the Euler line of triangle A'B'C'.

(1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
at one point on the nine-point circle of triangle ABC.

Note that (1a) follows from the definition of the Schiffler
point (H is the incenter of triangle A'B'C'), and (1b) is a
result of Victor Thebault. Both Euler lines of triangles
HB'C' and AB'C' pass through the midpoint Ha of AH, since
this midpoint is the circumcenter of both triangles. If Hb
and Hc are the midpoints of BH and CH, we have analogous
results. On the other hand, it is well-known that the
points Ha, Hb, Hc lie on the nine-point circle of triangle
ABC and the triangle HaHbHc is homothetic to triangle ABC.

Jean-Pierre Ehrmann gave the following results in
Hyacinthos messages #3693 and #3694:

(2a) The Euler lines of triangles A'AbAc, B'BcBa, C'CaCb
concur at one point, namely the
complement-of-the-Schiffler point of triangle
A'B'C'. It lies on the Euler line of triangle A'B'C'.

(2b) The Euler lines of triangles AAbAc, BBcBa, CCaCb concur
at one point.

The point of concurrence in (2a) is called FIRST EHRMANN POINT
of triangle ABC, included as X(973) in Kimberling's ETC, and
the point of concurrence in (2b) is called SECOND EHRMANN
POINT, included as X(974).

Some days ago, I have found synthetic proofs of (2b) and a
part of (2a). We begin with (2b):

Let Ga, Gb, Gc be the centroids of triangles AAbAc, BBcBa,
CCaCb. Since CbAb || CA, we have

d(Cb; CA) = d(Ab; CA).

[Here, d(P; g) means the distance from a point P to a line g.]

Since the centroid of a triangle trisects the medians, we
have

d(Gc; CA) = d(Cb; CA)/3
and d(Ga; CA) = d(Ab; CA)/3,

hence d(Gc; CA) = d(Ga, CA), and GcGa || CA. Analogously,
GaGb || AB and GbGc || BC. Hence, triangles GaGbGc and ABC
are homothetic. Since triangle ABC is homothetic with triangle
HaHbHc, triangles GaGbGc and HaHbHc are homothetic. But the
Euler lines of triangles AAbAc, BBcBa, CCaCb pass through Ga,
Gb, Gc, respectively (the centroid of a triangle lies on its
Euler line!), while the Euler lines of triangles AB'C', BC'A',
CA'B' pass through Ha, Hb, Hc, respectively. These Euler
lines are respectively parallel (since triangles AB'C' and
AAcAb are homothetic, and similarly for B and C). Hence, from
the concurrence of the Euler lines of triangles AB'C', BC'A',
CA'B' at a point on the nine-point circle of triangle ABC
(i. e., on the circumcircle of triangle HaHbHc), we conclude
the concurrence of the Euler lines of triangles AAbAc, BBcBa,
CCaCb on the circumcircle of triangle GaGbGc. Herewith, we
have proven even more than (2b)!

Now to (2a).

Call Wa, Wb, Wc the orthocenters of triangles A'AbAc, B'BcBa,
C'CaCb. Since A'Ab is orthogonal to AB, while AcBc is parallel
to AB, the line AcBc is an altitude of triangle A'AbAc and
passes through its orthocenter Wa. Analogously, AcBc passes
through Wb. Hence, the line WaWb coincides with AcBc and is
therefore parallel to AB. Analogously, WbWc is parallel to
BC and WcWa is parallel to CA. Hence, triangles WaWbWc and
ABC are homothetic. Since triangles HaHbHc and ABC are also
homothetic, we get the homothety of triangles WaWbWc and
HaHbHc. But the Euler lines of triangles A'AbAc, B'BcBa,
C'CaCb pass through Wa, Wb, Wc, respectively (the orthocenter
of a triangle lies on its Euler line!), while the Euler
lines of triangles HB'C', HC'A', HA'B' pass through Ha, Hb,
Hc, respectively. These Euler lines are respectively
parallel (since triangles HB'C' and A'AcAb are homothetic,
and analogously for B and C). Hence, from the concurrence
of the Euler lines of triangles HB'C', HC'A', HA'B', we
deduce the concurrence of the Euler lines of triangles
A'AbAc, B'BcBa, C'CaCb. This yields a part of (2a).

Maybe, using some special properties of the configuration
(Taylor hexagon of triangle ABC and "Spieker hexagon" of
triangle A'B'C') we can obtain a proof of the rest of (2a).

Darij Grinberg
• Dear Darij in Hyacinthos 7408, you wrote ... Note that if an angle of ABC is obtuse, the incenter of A B C is not H, but the corresponding vertex of ABC. ...
Message 2 of 14 , Aug 1, 2003
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Dear Darij
in Hyacinthos 7408, you wrote
> Let ABC be a triangle and AA', BB', CC' its altitudes.
> We have the following theorems:
>
> (1a) The Euler lines of triangles HB'C', HC'A', HA'B' concur
> at one point, namely the Schiffler point of triangle
> A'B'C'. It lies on the Euler line of triangle A'B'C'.
>
> (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
> at one point on the nine-point circle of triangle ABC.
>
> Note that (1a) follows from the definition of the Schiffler
> point (H is the incenter of triangle A'B'C')

Note that if an angle of ABC is obtuse, the incenter of A'B'C' is
not H, but the corresponding vertex of ABC.

> and (1b) is a result of Victor Thebault.
Notethat the common point is the center of Jerabek hyperbola.
Friendly. Jean-Pierre
• Dear Jean-Pierre Ehrmann, ... Yes, of course I know this, the excenters of a triangle also lie on the Neuberg cubic. ... I have met this before. The point M
Message 3 of 14 , Aug 1, 2003
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Dear Jean-Pierre Ehrmann,

Thanks for the mail. You wrote:

>> > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
>> > at one point on the nine-point circle of triangle ABC.
>> >
>> > Note that (1a) follows from the definition of the Schiffler
>> > point (H is the incenter of triangle A'B'C')
>>
>> Note that if an angle of ABC is obtuse, the incenter of A'B'C' is
>> not H, but the corresponding vertex of ABC.

Yes, of course I know this, the excenters of a triangle
also lie on the Neuberg cubic.

>> > and (1b) is a result of Victor Thebault.
>> Notethat the common point is the center of Jerabek hyperbola.

I have met this before. The point M where the Euler lines
of triangles AB'C', BC'A', CA'B' meet has the property
that one of the equations MA' = MB' + MC' or cyclically
holds. This was stated by Thebault. Can anybody find a
SYNTHETIC proof?

Thanks,
Sincerely,
Darij Grinberg
• Dear Darij [DG] ... [JPE] ... is ... [DG] ... I was just meaning that the common point is the Schiffler point of A B C only when ABC is acutangle ... I m
Message 4 of 14 , Aug 2, 2003
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Dear Darij

[DG]
> >> > (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
> >> > at one point on the nine-point circle of triangle ABC.
> >> >
> >> > Note that (1a) follows from the definition of the Schiffler
> >> > point (H is the incenter of triangle A'B'C')

[JPE]
> >> Note that if an angle of ABC is obtuse, the incenter of A'B'C'
is
> >> not H, but the corresponding vertex of ABC.

[DG]
> Yes, of course I know this, the excenters of a triangle
> also lie on the Neuberg cubic.

I was just meaning that the common point is the Schiffler point of
A'B'C' only when ABC is acutangle

> >> > and (1b) is a result of Victor Thebault.
> >> Notethat the common point is the center of Jerabek hyperbola.

> I have met this before. The point M where the Euler lines
> of triangles AB'C', BC'A', CA'B' meet has the property
> that one of the equations MA' = MB' + MC' or cyclically
> holds. This was stated by Thebault. Can anybody find a
> SYNTHETIC proof?

I'm unable to find a synthetic proof of this fact but it is quite
easy to check that
MA' : MB' : MC' = |(b^2-c^2)SA| : |(c^2-a^2)SB| : |(a^2-b^2)SC| and
(b^2-c^2)SA+(c^2-a^2)SB+(a^2-b^2)SC = 0
Friendly. Jean-Pierre
• Dear all, ... I think that a synthetic proof (in Russian) can be found at http://archive.1september.ru/mat/2000/no43_1.htm Alas, I cannot follow the author s
Message 5 of 14 , Aug 3, 2003
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Dear all,

In Hyacinthos message #7410, I wrote:

>> >> > (1b) The Euler lines of triangles AB'C', BC'A',
>> >> > CA'B' concur at one point on the
>> >> > nine-point circle of triangle ABC.

>> >> > and (1b) is a result of Victor Thebault.
>> >> Notethat the common point is the center of
>> >> Jerabek hyperbola.

>> I have met this before. The point M where the Euler lines
>> of triangles AB'C', BC'A', CA'B' meet has the property
>> that one of the equations MA' = MB' + MC' or cyclically
>> holds. This was stated by Thebault. Can anybody find a
>> SYNTHETIC proof?

I think that a synthetic proof (in Russian) can be found at

http://archive.1september.ru/mat/2000/no43_1.htm

Alas, I cannot follow the author's observation. Anybody who
can help?

Darij Grinberg

PS: Thanks Jean-Pierre, perhaps your equations can be
helpful in finding a synthetic proof.

Darij
• ... This was once discussed here (Floor had proposed it as a problem in the AMM), and were generalizations by PY and APH I don t remember in what direction,
Message 6 of 14 , Sep 11, 2003
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in Hyacinthos 7408 Darij wrote:

> Let ABC be a triangle and AA', BB', CC' its altitudes.
> We have the following theorems:
> (1b) The Euler lines of triangles AB'C', BC'A', CA'B' concur
> at one point on the nine-point circle of triangle ABC.

This was once discussed here (Floor had proposed it
as a problem in the AMM), and were generalizations
by PY and APH

I don't remember in what direction, but now let's
parametrize it.

Let ABC be a triangle and AHa, BHb, CHc its altitudes.

Special case #1:

Let Ab, Ac be the reflections of A in Hc, Hb, resp.
Let Bc, Ba be the reflections of B in Ha, Hc, resp.
Let Ca, Cb be the reflections of C in Hb, Ha, resp.

The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
concur (at H of ABC).

Problem: Are their Nine Point Circles concurrent?

Special Case #2:

Let Mac, Mab be the midpoints of AHc, AHb, resp.
Let Mba, Mbc be the midpoints of BHa, BHc, resp.
Let Mcb, Mca be the midpoints of CHb, CHa, resp.

The Euler Lines of the Triangles AMabMac, BMbcMba,
CMcaMcb are concurrent (??)

Generalization:

Let Pac, Pab be points on AB, AC resp. such that:

APac / AHc = APab / AHb = t

Similarly we define the points Pba, Pbc; Pcb, Pca

The Euler Lines of the Triangles APabPac, BPbcPba,
CPcaPcb are concurrent (??) Locus??

APH

--
• Dear Antreas, ... ^^^^^^^^^^^^^^^^^^^ I guess you mean AAbAc, BBcBa, CCaCb. The Euler lines of these triangles concur since the orthocenter
Message 7 of 14 , Sep 11, 2003
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Dear Antreas,

In Hyacinthos message #7858, you wrote (partly):

>> Let ABC be a triangle and AHa, BHb, CHc its altitudes.
>>
>> Special case #1:
>>
>> Let Ab, Ac be the reflections of A in Hc, Hb, resp.
>> Let Bc, Ba be the reflections of B in Ha, Hc, resp.
>> Let Ca, Cb be the reflections of C in Hb, Ha, resp.
>>
>> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
^^^^^^^^^^^^^^^^^^^
I guess you mean AAbAc, BBcBa, CCaCb.
The Euler lines of these triangles concur since the
orthocenter H of triangle ABC is their common
circumcenter.

>> concur (at H of ABC).

>> Problem: Are their Nine Point Circles concurrent?

Yes. In fact, the nine-point circle of a triangle
ABC is the reflection of the circumcircle of triangle
AB'C' in the line B'C', where A', B', C' are the
midpoints of the sides BC, CA, AB. This yields that
the nine-point circle of triangle AAbAc is the
reflection of the circumcircle of triangle AHbHc in
the line HbHc. Similarly for the other two nine-point
circles. Since the circles AHbHc, BHcHa, CHaHb concur
at H, the three nine-point circles concur at the
antigonal conjugate of H with respect to triangle
HaHbHc.

NOTE. Antigonal conjugates? If P is a point in the
plane of a triangle ABC, then the reflections of the
circles PBC, PCA, PAB in the sidelines BC, CA, AB
have a common point. This point is called antigonal
conjugate of P with respect to triangle ABC. But I
see (Hyacinthos message #7826) that you know this,
maybe with another name.

The point where your nine-point circles concur is
not in the ETC.

>> Generalization:
>>
>> Let Pac, Pab be points on AB, AC resp. such that:
>>
>> APac / AHc = APab / AHb = t
>>
>> Similarly we define the points Pba, Pbc; Pcb, Pca
>>
>> The Euler Lines of the Triangles APabPac, BPbcPba,
>> CPcaPcb are concurrent

Yes, they are: the proof is simple. Moreover, if
J = X(125) is the intersection of the Euler lines of
triangles AHbHc, BHcHa, CHaHb, then the Euler lines
of triangles APabPac, BPbcPba, CPcaPcb meet at the
point W lying on the line HJ and dividing the segment
HJ in the ratio

HW 2-t
-- = ---.
WJ t-1

Remark that the line HJ passes through X(74).

Sincerely,
Darij Grinberg
• Dear Darij [APH, carelessly!] ... I don t know if Paul and I have already discussed this point in the past (we had discussed a number of cases of concurrent
Message 8 of 14 , Sep 11, 2003
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Dear Darij

[APH, carelessly!]
>>> Let ABC be a triangle and AHa, BHb, CHc its altitudes.
>>>
>>> Special case #1:
>>>
>>> Let Ab, Ac be the reflections of A in Hc, Hb, resp.
>>> Let Bc, Ba be the reflections of B in Ha, Hc, resp.
>>> Let Ca, Cb be the reflections of C in Hb, Ha, resp.
>>>
>>> The Euler Lines of the Triangles AHbHc, BHcHa, CHaHb
> ^^^^^^^^^^^^^^^^^^^

[DG]:

>I guess you mean AAbAc, BBcBa, CCaCb.
>The Euler lines of these triangles concur since the
>orthocenter H of triangle ABC is their common
>circumcenter.

[APH]:
>>> concur (at H of ABC).
>
>>> Problem: Are their Nine Point Circles concurrent?

[DG]:
>Yes. In fact, the nine-point circle of a triangle
>ABC is the reflection of the circumcircle of triangle
>AB'C' in the line B'C', where A', B', C' are the
>midpoints of the sides BC, CA, AB. This yields that
>the nine-point circle of triangle AAbAc is the
>reflection of the circumcircle of triangle AHbHc in
>the line HbHc. Similarly for the other two nine-point
>circles. Since the circles AHbHc, BHcHa, CHaHb concur
>at H, the three nine-point circles concur at the
>antigonal conjugate of H with respect to triangle
>HaHbHc.
>
>NOTE. Antigonal conjugates? If P is a point in the
>plane of a triangle ABC, then the reflections of the
>circles PBC, PCA, PAB in the sidelines BC, CA, AB
>have a common point. This point is called antigonal
>conjugate of P with respect to triangle ABC. But I
>see (Hyacinthos message #7826) that you know this,
>maybe with another name.
>
>The point where your nine-point circles concur is
>not in the ETC.

I don't know if Paul and I have already discussed this point
in the past (we had discussed a number of cases of
concurrent Nine point Circles)
Is it in your archive, Paul?

Anyway, it is a remarkable point and has place
in Kimberling's ETC, but what its coordinates are?

Of course, the locus problem is:

Which is the locus of the radical center of
the nine point circles in the parametrization below,
but I don't think that the curve is something
interesting.

[APH]:
>>> Generalization:
>>>
>>> Let Pac, Pab be points on AB, AC resp. such that:
>>>
>>> APac / AHc = APab / AHb = t
>>>
>>> Similarly we define the points Pba, Pbc; Pcb, Pca
>>>
>>> The Euler Lines of the Triangles APabPac, BPbcPba,
>>> CPcaPcb are concurrent

[DG]:

>Yes, they are: the proof is simple. Moreover, if
>J = X(125) is the intersection of the Euler lines of
>triangles AHbHc, BHcHa, CHaHb, then the Euler lines
>of triangles APabPac, BPbcPba, CPcaPcb meet at the
>point W lying on the line HJ and dividing the segment
>HJ in the ratio
>
> HW 2-t
> -- = ---.
> WJ t-1
>
>Remark that the line HJ passes through X(74).

Thanks

Greetings

Antreas
--
• Let ABC be a triangle, HaHbHc its orthic triangle, and EaEbEc its Euler triangle (ie Ea = midpoint of AH etc). HbEa / AB = Hab HcEa / AC = Hac Similarly the
Message 9 of 14 , Sep 11, 2003
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Let ABC be a triangle, HaHbHc its orthic triangle, and
EaEbEc its Euler triangle (ie Ea = midpoint of AH etc).

HbEa /\ AB = Hab

HcEa /\ AC = Hac

Similarly the points Hbc, Hba; Hca, Hcb.

Are the Euler Lines of the Triangles AHabHac,
BHbcHba, CHcaHcb concurrent ?

Parametrization:

Ea, Eb, Ec points on AH, BH, CH, resp.
such that:

AEa / AH = BEb / BH = CEc / CH = t.

etc

Antreas

--
• ... Barycentrics (sin 3A)(1 + cos 2B + cos 2C)/(cos A) : ... : ... It was a routine computer calculation to get an answer. Reducing that answer to this short
Message 10 of 14 , Sep 12, 2003
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[APH]:
> Anyway, it is a remarkable point and has place
> in Kimberling's ETC, but what its coordinates are?

Barycentrics
(sin 3A)(1 + cos 2B + cos 2C)/(cos A) : ... : ...

It was a routine computer calculation to get an answer. Reducing
that answer to this short formula was another problem.
--
Barry Wolk

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• Dear colleagues! What is the number in ETC of Thebault point, i.e the common point of euler lines of triangles AB C , A BC and A B C where AA , BB , CC are
Message 11 of 14 , Dec 2, 2007
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Dear colleagues!
What is the number in ETC of Thebault point, i.e the common point of euler lines of triangles AB'C', A'BC' and A'B'C where AA', BB', CC' are the altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this point is on Steiner inellipse . Is this correct?

Sincerely Alexey

Antivirus scanning: Symantec Mail Security for SMTP.

[Non-text portions of this message have been removed]
• Dear Alexey ... euler lines of triangles AB C , A BC and A B C where AA , BB , CC are the altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this
Message 12 of 14 , Dec 3, 2007
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Dear Alexey

> What is the number in ETC of Thebault point, i.e the common point of
euler lines of triangles AB'C', A'BC' and A'B'C where AA', BB', CC' are
the altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this
point is on Steiner inellipse . Is this correct?

I think that your point is the center of the Jerabek hyperbola (X_125)
As the four common points of the Steiner inellipse with the 9Pcircle
are the midpoints of the sides and the center of the Kiepert hyperbola,
I don't think that this point lies on the Steiner inellipse
Friendly. Jean-Pierre
• Dear Alexey, [AZ]: What is the number in ETC of Thebault point, i.e the common point of euler lines of triangles AB C , A BC and A B C where AA , BB , CC are
Message 13 of 14 , Dec 3, 2007
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Dear Alexey,

[AZ]: What is the number in ETC of Thebault point, i.e the common
point of euler lines of triangles AB'C', A'BC' and A'B'C where AA',
BB', CC' are the altitudes of ABC? Moscow scholar Fyodor Nilov asserts
that this point is on Steiner inellipse . Is this correct?

*** The Euler lines of the three triangles intersect at the center of
the Jerabek hyperbola. It is the point with barycentric coordinates

((b^2+c^2-a^2)(b^2-c^2)^2 : ... : ...).

In ETC, this is X(125), called the Jerabek center.

It does not lie on the Steiner inellipse.

Best regards
Sincerely
Paul
• Dear Alexey Let T be the Thebault point and (L) the Euler line of ABC. The lines symmetric of (L) wrt the sides BC, CA, AB are on a same point U on the
Message 14 of 14 , Dec 3, 2007
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Dear Alexey
Let T be the Thebault point and (L) the Euler line of ABC.
The lines symmetric of (L) wrt the sides BC, CA, AB are on a same point U on
the ABC-circumcircle.
Then T is homothetic of U in the homothety of center G, the ABC-centroid and
ratio -1/2.
I think T is not on the ABC-Steiner inellipse.
Friendly
Francois

On Dec 3, 2007 8:24 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

> Dear colleagues!
> What is the number in ETC of Thebault point, i.e the common point of euler
> lines of triangles AB'C', A'BC' and A'B'C where AA', BB', CC' are the
> altitudes of ABC? Moscow scholar Fyodor Nilov asserts that this point is on
> Steiner inellipse . Is this correct?
>
> Sincerely Alexey
>
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>

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