## Fuhrmann triangle

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• Let ABC be a triangle, and A , B , C be the midpoints of the arcs BC, CA, AB on the circumcircle of ABC. Denote by A , B , C the reflections of A , B , C in
Message 1 of 7 , Jan 9, 2003
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Let ABC be a triangle, and A', B', C' be the midpoints of the arcs
BC, CA, AB on the circumcircle of ABC. Denote by A", B", C" the
reflections of A', B', C' in BC, CA, AB. Then the triangle A"B"C" is
called FUHRMANN TRIANGLE of ABC.

It is well-known that the circumcenter of A"B"C" is the midpoint M of
segment HN, where H is the orthocenter and N the Nagel point of
triangle ABC. One can easily prove that this midpoint M is also the
reflection of the incenter I of ABC in the nine-point center O', and
also the reflection of the circumcenter O in the Spieker center I'
(what an analogy between incenter and circumcenter!).

Something very interesting was stated by Milorad Stevanovich in
Hyacinthos message #5897: The orthocenter of the Fuhrmann triangle
A"B"C" is the incenter of I of ABC.

This is, as you can easily see, equivalent to the following fact: The
nine-point centers of ABC and the Fuhrmann triangle A"B"C" coincide.

However, I do not know a proof of this theorem either in Milorad's or
in my form.

Another conjecture which I have found:

The perpendiculars from A to B"C", from B to C"A", and from C to A"B"
concur at the point X(80), namely the reflection of the incenter I in
the Feuerbach point of ABC.

(The fact that they concur follows from the orthologicity of
triangles ABC and A"B"C", but I suppose it should be not so easy to
prove that the concurrence is X(80).)

-- I could keep on finding new and new interesting properties of the
Fuhrmann triangle, but I know that nobody will read a long message;
so I conclude with only one little note: The Fuhrmann triangle has
some properties similar to those of the two Brocard triangles,
although it is not perspective to ABC. Altogether, the Fuhrmann
triangle should be much better known as it is.

Darij Grinberg
• ... is ... of ... and ... This isn t strictly about the Fuhrmann triangle, but it s related to it. I was playing around in Euklides trying to show that H is on
Message 2 of 7 , Apr 18, 2004
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--- In Hyacinthos@yahoogroups.com, "Darij Grinberg
<darij_grinberg@w...>" <darij_grinberg@w...> wrote:
> Let ABC be a triangle, and A', B', C' be the midpoints of the arcs
> BC, CA, AB on the circumcircle of ABC. Denote by A", B", C" the
> reflections of A', B', C' in BC, CA, AB. Then the triangle A"B"C"
is
> called FUHRMANN TRIANGLE of ABC.
>
> It is well-known that the circumcenter of A"B"C" is the midpoint M
of
> segment HN, where H is the orthocenter and N the Nagel point of
> triangle ABC. One can easily prove that this midpoint M is also the
> reflection of the incenter I of ABC in the nine-point center O',
and
> also the reflection of the circumcenter O in the Spieker center I'
> (what an analogy between incenter and circumcenter!).

This isn't strictly about the Fuhrmann triangle, but it's related to
it. I was playing around in Euklides trying to show that H is on the
Fhurmann circle, and I thought this property could be generalized:

ABC is a triangle and P is a point in its plane. AP, BP, CP cut the
circumcicle of ABC in A', B', C' respectively. A", B", C" are the
reflections of A', B', C" in BC, CA, AB respectively. Show that
A"B"C" is similar to A'B'C' (a) and H lies on the circumcircle of
A"B"C" (b).

I it stupid for me to mention this? :-) (I mean, is it really well-
known?). Anyway, I don't have a proof. Does ne1 have a nice proof (no
monstrous computations)?

It's easy to calculate the angles between HA", HB", HC", so it's
enough to give a proof for either (a) or (b) and the other one will
follow.

Thanks!
• Dear Grobber, ... This was mentioned before in Hyacinthos message #9493 by Jean-Pierre Ehrmann and proven (synthetically!) in the messages #9500 and #9503. A
Message 3 of 7 , Apr 18, 2004
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Dear Grobber,

In Hyacinthos message #9713, you wrote:

>> ABC is a triangle and P is a point in its plane.
>> AP, BP, CP cut the circumcicle of ABC in A', B',
>> C' respectively. A", B", C" are the reflections
>> of A', B', C" in BC, CA, AB respectively. Show
>> that A"B"C" is similar to A'B'C' (a) and H lies
>> on the circumcircle of A"B"C" (b).

This was mentioned before in Hyacinthos message
#9493 by Jean-Pierre Ehrmann and proven
(synthetically!) in the messages #9500 and #9503.
A reference to Hagge 1908 was given in message
#9506, and I gave some additional results in
#9511. By the way, I am still searching for
synthetic proofs of the conjectures (3) and (5)
from my message #9511. Actually, these two
conjectures can be united under

(6) There exists an indirect similitude mapping
the points A, B, C, A', B', C' and P to the
points X, Y, Z, A2, B2, C2 and P

(where the notations are taken from Hyacinthos
message #9511). I would be grateful for any
help regarding this conjecture.

Sincerely,
Darij Grinberg
• ... is ... of ... and ... The ... or ... A B ... in ... I ll make use of the fact (proven by Milorad Stevanovich) that I is the orthocenter of A B C and of
Message 4 of 7 , Apr 19, 2004
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--- In Hyacinthos@yahoogroups.com, "Darij Grinberg
<darij_grinberg@w...>" <darij_grinberg@w...> wrote:
> Let ABC be a triangle, and A', B', C' be the midpoints of the arcs
> BC, CA, AB on the circumcircle of ABC. Denote by A", B", C" the
> reflections of A', B', C' in BC, CA, AB. Then the triangle A"B"C"
is
> called FUHRMANN TRIANGLE of ABC.
>
> It is well-known that the circumcenter of A"B"C" is the midpoint M
of
> segment HN, where H is the orthocenter and N the Nagel point of
> triangle ABC. One can easily prove that this midpoint M is also the
> reflection of the incenter I of ABC in the nine-point center O',
and
> also the reflection of the circumcenter O in the Spieker center I'
> (what an analogy between incenter and circumcenter!).
>
> Something very interesting was stated by Milorad Stevanovich in
> Hyacinthos message #5897: The orthocenter of the Fuhrmann triangle
> A"B"C" is the incenter of I of ABC.
>
> This is, as you can easily see, equivalent to the following fact:
The
> nine-point centers of ABC and the Fuhrmann triangle A"B"C" coincide.
>
> However, I do not know a proof of this theorem either in Milorad's
or
> in my form.
>
> Another conjecture which I have found:
>
> The perpendiculars from A to B"C", from B to C"A", and from C to
A"B"
> concur at the point X(80), namely the reflection of the incenter I
in
> the Feuerbach point of ABC.

I'll make use of the fact (proven by Milorad Stevanovich) that I is
the orthocenter of A"B"C" and of the known fact that IO'=R/2-r, where
O' is the Euler center of ABC.

First of all, the centers of Fuhrmann's circle, of Euler's circle and
of the cirumcircle are collinear and I lies on this line, so X(80)
also lies on this line. Further on we can see that MI*MX(80)=(R-2r)(R-
2r+2r)=R^2-2Rr=OI^2=MH^2, so X(80) is the inverse of I with respect
to the Fuhrmann circle.

The last remark shows that triangles MIA" and MA"X(80) are similar,
so MI/MA"=IA"/A"X(80), but MI=R-2r and MA"=OI=sqrt(R^2-2Rr), so A"X
(80)=IA"*sqrt(R/(R-2r)) (*).

A"C=A'C=IA' and the system I, A", B", C" is similar to I, A', B', C'
(because A"B"C" is similar to A'B'C' as proven in some messages
around 9500-9511 for the general case and I is the orthocenter of
both A'B'C' and A"B"C"), so IA"/IA'=(circumradius of A"B"C")/R=sqrt
(R^2-2Rr)/R=>IA'=IA"*sqrt(R/(R-2r)) (**).

From (*) and (**) we find A"C=A"X(80). In the same way we show B"C=B"X
(80), so A"B" is the perpendicular bisector of CX(80), and from all
the analogous relations we find what we want: the perpendiculars from
A, B, C to B"C", C"A", A"B" concur at X(80).

(Here A', B', C' are the midpoints of the smaller arcs BC, CA, AB)
• Dear Grobber, ... Thanks - this remark of you about perpendicular bisectors has opened my eyes to see a much simpler proof: The main property of X(80) is the
Message 5 of 7 , Apr 19, 2004
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Dear Grobber,

In Hyacinthos message #9724, you wrote:

>> A"B" is the perpendicular bisector of CX(80),
>> and from all the analogous relations we find
>> what we want: the perpendiculars from A, B,
>> C to B"C", C"A", A"B" concur at X(80).

Thanks - this remark of you about perpendicular
bisectors has opened my eyes to see a much simpler
proof:

The main property of X(80) is the following: If
the points Ja, Jb, Jc are the reflections of the
excenters Ia, Ib, Ic of triangle ABC in the
sidelines BC, CA, AB, then the lines AJa, BJb, CJc
and the circles BCJa, CAJb, ABJc all pass through
X(80). Now, it follows that the circle BCX(80)
coincides with the circle BCJa. Hence, the
circumcenter of triangle BCX(80) is the
circumcenter of triangle BCJa. But since the
triangle BCJa is the reflection of the triangle
BCIa in the line BC, the circumcenter of triangle
BCJa is the reflection of the circumcenter of
triangle BCIa in BC, i. e. the reflection of A' in
BC (since A' is the circumcenter of triangle BCIa,
what is very easy to prove), i. e. the point A".
In other words, the point A" is the circumcenter
of triangle BCX(80). Similarly, the points B" and
C" are the circumcenters of triangles CAX(80) and
ABX(80). Consequently, the lines B"C", C"A", A"B"
are the perpendicular bisectors of the segments
AX(80), BX(80), CX(80). Qed..

Now you may ask: How to prove that the lines AJa,
BJb, CJc and the circles BCJa, CAJb, ABJc all pass
through the point X(80) which is initially defined
as the reflection of the incenter I in the
Feuerbach point of triangle ABC ?

Well, this is quite tricky. At first, you consider
four arbitrary points A, B, C, D in the plane, and
show by some angle chasing that the nine-point
circles of the triangles ABC, BCD, CDA, DAB concur
at one point P. We call this point P the "Poncelet
point" of the points A, B, C, D. This Poncelet
point P also lies on the pedal circles of the
points A, B, C, D with respect to the triangles
BCD, CDA, DAB, ABC. [The "pedal circle" of a point
with respect to a triangle is defined as the
circle passing through the orthogonal projections
of the point on the sidelines of the triangle.]
Prove this by angle chase again (this time, you
must use the feet of the altitudes of the
triangles BCD, CDA, DAB, ABC; these feet lie on
the nine-point circles and on the appropriate
pedal circles as well!). Now let D' be the
reflection of D in the Poncelet point P. Then you
will easily show (angle chase again) that
< BD'C = - < BDC, < CD'A = - < CDA and
< AD'B = - < ADB. This point D' is called the
"antigonal conjugate" of the point D with respect
to triangle ABC; the word "antigonal" means
nothing but the fact that the angles at which we
see the triangle's sides from this point are
oppositely equal to the corresponding angles for
D (i. e. the fact that < BD'C = - < BDC, etc.).
Antigonal conjugate is NOT the same as isogonal
conjugate. (In fact, these two kinds of
conjugation are interrelated: One can prove that
the antigonal conjugate D' of D is the inverse of
the isogonal conjugate of the inverse of D;
hereby, the word "inverse" of a point always means
the image of the point in the inversion with
respect to the circumcircle of triangle ABC. But
we won't need this result in our proof.)

Okay, well, we have introduced the notions of the
Poncelet point and the antigonal point. Now, let D
be the incenter I of triangle ABC. What is the
Poncelet point P of the points A, B, C, D = I ?
Well, the Poncelet point P must lie on the
nine-point circles of triangles ABC, BCD, CDA, DAB
and on the pedal circles of the points A, B, C, D
with respect to the triangles BCD, CDA, DAB, ABC;
however, if you consider only two of these eight
circles, namely the nine-point circle of triangle
ABC and the pedal circle of the point D = I with
respect to triangle ABC, then already these two
circles have only one common point, namely the
Feuerbach point of triangle ABC (this is just the
great Feuerbach theorem). Hence, the Poncelet
point P is the Feuerbach point of triangle ABC.
Now, you consider the antigonal conjugate D' of D
with respect to triangle ABC. By definition, D' is
the reflection of the incenter I in the Feuerbach
point P; hence, D' is X(80). But from the equatins
< BD'C = - < BDC, < CD'A = - < CDA and
< AD'B = - < ADB, we can derive by a simple angle
chase that the point D' lies on the circles BCJa,
CAJb, ABJc, and in the next step we can easily
conclude by another bit of angle calculation that
this point D' also lies on the lines AJa, BJb,
CJc. Proof complete!

This was quite much theory now; feel free to ask
me if you don't understand something, I have
omitted many details. Also, what I have written is
only a little part of the theory around Poncelet
points and antigonal conjugates. If you are
interested, I can tell you more of the story.

Sincerely,
Darij Grinberg
• Dear Darij, in message 9725 you wrote about the antigonal conjugate: ... ... I think it is: The antigonal conjugate D of D is the isogonal conjugate of the
Message 6 of 7 , May 4 11:15 AM
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Dear Darij,

in message 9725 you wrote about the antigonal conjugate:

....
> One can prove that
> the antigonal conjugate D' of D is the inverse of
> the isogonal conjugate of the inverse of D;
> hereby, the word "inverse" of a point always means
> the image of the point in the inversion with
> respect to the circumcircle of triangle ABC.
...

I think it is:

The antigonal conjugate D' of D is the isogonal conjugate of
the inverse of the isogonal conjugate of D

I suppose you've interchanged the terms inverse and isogonal
conjugate ?

Greetings from Bruges

Eric Danneels
• Dear Eric, ... Oh yes, sorry. Thanks for the remark. By the way, the proof is easy: If D1 is the isogonal conjugate of D, then we can readily see that
Message 7 of 7 , May 5 12:16 PM
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Dear Eric,

In Hyacinthos message #9761, you wrote:

>> I think it is:
>>
>> The antigonal conjugate D' of D is the
>> isogonal conjugate of the inverse of the
>> isogonal conjugate of D
>>
>> I suppose you've interchanged the terms
>> inverse and isogonal conjugate ?

Oh yes, sorry. Thanks for the remark. By
the way, the proof is easy: If D1 is the
isogonal conjugate of D, then we can
readily see that < BD1C = < BAC - < BDC.
If D2 is the inverse of D1, then another
angle chase shows < BD2C = 2 < BAC - < BDC,
so that < BD2C = < BAC + < BDC. Now, if
D3 is the isogonal conjugate of D2, then
< BD3C = < BAC - < BD2C, so that
< BD3C = - < BDC = < BD'C, and similarly
< CD3A = < CD'A and < AD3B = < AD'B, and
consequently, D3 coincides with the
antigonal conjugate D' of D.

Sincerely,
Darij Grinberg
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