- 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 - Dear Eric,

In Hyacinthos message #9761, you wrote:

>> I think it is:

Oh yes, sorry. Thanks for the remark. By

>>

>> 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 ?

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