## Re: [EMHL] Fuhrmann

Expand Messages
• Saturday, January 15, 2005 9:53 PM [GMT+1=CET], ... In Episodes in Nineteenth and Twentieth Cantury Euclidean Geometry (chapter six: The Fuhrmann Circle ),
Message 1 of 13 , Jan 16, 2005
• 0 Attachment
Saturday, January 15, 2005 9:53 PM [GMT+1=CET],

> Sorry the following I said
>> The following site
>>
> http://felix.unife.it/Root/d-Mathematics/d-The-mathematician/t-Mathematicians-A-Z#F
>>
>> gives that he was a Danish mathematician
>
> that is a Danish mathematician is not correct.
>

In "Episodes in Nineteenth and Twentieth Cantury Euclidean Geometry"
(chapter six: "The Fuhrmann Circle"), Ross Honsberger saids that Wilhelm
Fuhrmann was a 19th-century German geometer (1833-1904).

Saludos,

Ignacio Larrosa Canestro
A Coruna (Espana)
ilarrosa@...
• Dear friends, If ABC is a triangle and P = f(n) is an interior point with barycnetris (AP^n : BP^n : CP^n) then it is obvious that f(0) = G the centroid of
Message 2 of 13 , Jan 16, 2005
• 0 Attachment
Dear friends,

If ABC is a triangle and P = f(n) is an interior point
with barycnetris (AP^n : BP^n : CP^n)
then it is obvious that f(0) = G the centroid of ABC.
Which is the point f(-1)?
Can we find the points f(1), f(2), f(-2)?

Best regards
• Dear Nikolaos ... Some of them have been discussed in Haycinthos (note that sometimes there are many solutions - or no solution - for a given n) n = -1
Message 3 of 13 , Jan 16, 2005
• 0 Attachment
Dear Nikolaos
> If ABC is a triangle and P = f(n) is an interior point
> with barycnetris (AP^n : BP^n : CP^n)
> then it is obvious that f(0) = G the centroid of ABC.
> Which is the point f(-1)?
> Can we find the points f(1), f(2), f(-2)?

Some of them have been discussed in Haycinthos (note that sometimes
there are many solutions - or no solution - for a given n)
n = -1
Vect(AM)/AM +... = 0 gives the first Fermat point if the angles of
A,B,C are <120, nothing elsewhere.
n= -2
If z,za,zb,zc are the complex affixes of P,A,B,C, this means that
1/(z-a)+1/(z-b)+1/(z-c)=0. Hence two solutions : the focii of the
Steiner inellipse
n = 2
The solutions are lying on the Kiepert hyperbola of the medial
triangle.
More precisely : if we consider the circular circumcubic through X
(13),X(14),X(15),X(16),X(39),X(76),X(538),X(755) (Bernard???) the
solutions are exactly the common points - apart X(39) - of the cubic
and the hyperbola who are inside ABC. Hence, there exists at most 5
solutions. Of course, a proof neads a very awful computation and, in
any case, I don't know the number of solutions.
n = 1 seems very complicated
Friendly. Jean-Pierre
• Dear Nikolaos and Jean-Pierre, ... What a coincidence ! I ve been working on this subject these latter days for n = +/-2. The cubic is K290 = O(X39)
Message 4 of 13 , Jan 16, 2005
• 0 Attachment
Dear Nikolaos and Jean-Pierre,

> [ND] If ABC is a triangle and P = f(n) is an interior point
> > with  barycnetris (AP^n : BP^n : CP^n)
> > then it is obvious that f(0) = G the centroid of ABC.
> > Which is the point f(-1)?
> > Can we find the points f(1), f(2), f(-2)?
>
> [snip]
> n = 2
> The solutions are lying on the Kiepert hyperbola of the medial
> triangle.
> More precisely : if we consider the circular circumcubic through X
> (13),X(14),X(15),X(16),X(39),X(76),X(538),X(755) (Bernard???) the
> solutions are exactly the common points - apart X(39) - of the cubic
> and the hyperbola who are inside ABC. Hence, there exists at most 5
> solutions. Of course, a proof neads a very awful computation and, in
> any case, I don't know the number of solutions.

What a coincidence ! I've been working on this subject these latter
days for n = +/-2.

The cubic is K290 = O(X39) (orthopivotal cubic).
I have proven that the solutions are inside the in-ellipse with
perspector X(76).

I conjecture that there is one real solution and 4 imaginary.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear friends, here is a property of X(37), for which a would appreciate any references. Construct the minimal in perimeter/area equilateral hexagon, inscribed
Message 5 of 13 , Jan 17, 2005
• 0 Attachment
Dear friends,
here is a property of X(37), for which a would appreciate any references.

Construct the minimal in perimeter/area equilateral hexagon, inscribed in a
triangle ABC and having two vertices on each side. The minimal hexagon is
symmetric. Its sides are either parts of, or parallel to, the sides of the
triangle and its center of symmetry coincides with the triangle-center
X(37).

Look at
html
for a related picture

Another related question is about the inscribed conic (with center at x(37))
is it well known?
Do we know some other properties of this conic?
• Dear Nikolaos and Bernard [ND] If ABC is a triangle and P = f(n) is an interior point ... [JPE] ... through X ... the ... cubic ... most 5 ... and, in ... [BG]
Message 6 of 13 , Jan 17, 2005
• 0 Attachment
Dear Nikolaos and Bernard
[ND] If ABC is a triangle and P = f(n) is an interior point
> > > with  barycnetris (AP^n : BP^n : CP^n)
> > > then it is obvious that f(0) = G the centroid of ABC.
> > > Which is the point f(-1)?
> > > Can we find the points f(1), f(2), f(-2)?
[JPE]
> > n = 2
> > The solutions are lying on the Kiepert hyperbola of the medial
> > triangle.
> > More precisely : if we consider the circular circumcubic
through X
> > (13),X(14),X(15),X(16),X(39),X(76),X(538),X(755) (Bernard???)
the
> > solutions are exactly the common points - apart X(39) - of the
cubic
> > and the hyperbola who are inside ABC. Hence, there exists at
most 5
> > solutions. Of course, a proof neads a very awful computation
and, in
> > any case, I don't know the number of solutions.
[BG]
> The cubic is K290 = O(X39) (orthopivotal cubic).
> I have proven that the solutions are inside the in-ellipse with
> perspector X(76).
>
> I conjecture that there is one real solution and 4 imaginary.

If P = x:y:z, let Fa(P) = zBP^2-yCP^2; Fb(P) = xCP^2-zAP^2;
Fc(P)= yAP^2-xBP^2
Then Fa(P)=0, Fb(P)=0; Fc(P)=0 are three circular cubics and they
have 5 other common points apart the circular points at infinity.
These points are the required points.
Now Fa(P)+Fb(P)+Fc(P) = 0 is the Kiepert hyperbola of the medial
triangle and
Fa(P)/a^2+Fb(P)/b^2+Fc(P)/c^2=0 is K290 (thanks Bernard)
The cubic and the hyperbola intersect at X(39); their 5 other common
points are the required points
Using the parametrization x=1/(SB+t)+1/(SC+t), ...cyclic of the
hyperbola, we get a ugly equation of degree 5 for t. The problem is
to find the number of real roots (at least one, of course) of the
equation.
Note that, as the Galois group of the equation is S(5), our 5 points
are not even conic constructible thus there is no easier way to
construct these points than the way above.
Friendly. Jean-Pierre
• Dear friends, ... Hence we have the following construction of X(37). We extend the sides AB, AC to BC , CB such that BC =BC=CB = a. From C we draw C B ||=
Message 7 of 13 , Jan 17, 2005
• 0 Attachment
Dear friends,

--- Paris Pamfilos <pamfilos@...> wrote:
>
> Dear friends,
> here is a property of X(37), for which a would
> appreciate any references.
>
> Construct the minimal in perimeter/area equilateral
> hexagon, inscribed in a
> triangle ABC and having two vertices on each side.
> The minimal hexagon is
> symmetric. Its sides are either parts of, or
> parallel to, the sides of the
> triangle and its center of symmetry coincides with
> the triangle-center
> X(37).

Hence we have the following construction of X(37).
We extend the sides AB, AC to BC', CB' such that
BC'=BC=CB'= a.
From C' we draw C'B" ||= CB'= a and
from B' we draw B'C" ||= BC'= a
The segments BC", CB", B'C' have the same midpoint A*.
Similarly define the points B*, C*.
The distance of B from AC is asinC.
The distance of C" from AC is asinA
Hence the distance of A* from AC is a(sinA+sinC)/2.
Similarly
the distance of A* from AB is a(sinA+sinB)/2.
This means that the line AA* passes through the point
with trilinears (b+c : c+a : a+b) and this is X(37).
Hence the triangles ABC, A*B*C* are perspective
and the perspector is X(37).

Best regards

____________________________________________________________
Do You Yahoo!?
Αποκτήστε τη δωρεάν @... διεύθυνση σας στο http://www.otenet.gr
• We can also notice that the point of barycentrics (x(y+z),y(z+x),x(y+z)) is always the complement of the isotomic conjugate of the point of barycentrics
Message 8 of 13 , Jan 17, 2005
• 0 Attachment
We can also notice that the point
of barycentrics (x(y+z),y(z+x),x(y+z))
is always the complement of the isotomic conjugate
of the point of barycentrics (x,y,z), hence there is an affine
construction of the first point from the second,
here an affine construction of X(37) from the incenter X(1)
Best Regards
François Rideau
• ... Nikolaos s construction leads to many other perspectors. Using the following slightly changed notation: Let Ca be the point on AB such that BCa = BC (with
Message 9 of 13 , Jan 18, 2005
• 0 Attachment
> Hence we have the following construction of X(37).
> We extend the sides AB, AC to BC', CB' such that
> BC'=BC=CB'= a.
> From C' we draw C'B" ||= CB'= a and
> from B' we draw B'C" ||= BC'= a
> The segments BC", CB", B'C' have the same midpoint A*.
> Similarly define the points B*, C*.
> The distance of B from AC is asinC.
> The distance of C" from AC is asinA
> Hence the distance of A* from AC is a(sinA+sinC)/2.
> Similarly
> the distance of A* from AB is a(sinA+sinB)/2.
> This means that the line AA* passes through the point
> with trilinears (b+c : c+a : a+b) and this is X(37).
> Hence the triangles ABC, A*B*C* are perspective
> and the perspector is X(37).

Nikolaos's construction leads to many other perspectors.

Using the following slightly changed notation:

Let Ca be the point on AB such that BCa = BC
(with B between A and Ca)
Define Cb, Ab, Ac, Bc and Ba similarly
A*, B* and C* are the midpoints of CaBa, AbCb and BcAc
A1 is the intersection of AcBa and CaAb
Define B1 and C1 similarly
CaBa, AbCb and BcAc form a triangle A2B2C2
BcCb, AcCa and BaAb form a triangle A3B3C3
A4 is the intersection of CbAc and BcCa
Define B4 and C4 similarly

we have the following perspectivities

1) Triangle ABC and A*B*C* ==> Perspector X37
2) Triangle ABC and A1B1C1 ==> Perspector X8
3) Triangle ABC and A2B2C2 ==> Perspector X65
4) Triangle ABC and A3B3C3 ==> Perspector X65
5) Triangle A2B2C2 and A*B*C* ==> Perspector X210
6) Triangle A1B1C1 and A2B2C2 ==> Perspector not in ETC
7) Triangle A1B1C1 and A3B3C3 ==> Perspector not in ETC
8) Triangle ABC and A4B4C4 ==> Perspector X8
9) Triangle A2B2C2 and A4B4C4 ==> Perspector not in ETC
10) Triangle A3B3C3 and A4B4C4 ==> Perspector not in ETC
...

Best wishes for 2005

Eric

u(
Your message has been successfully submitted and would be delivered to recipients shortly.