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## Re: Proofs of Feuerbach theorem

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• ... So the nine point circle meets every such cevian circle. Does this characterise the nine-point circle? Or is there another circle that meets every cevian
Message 1 of 20 , Feb 29, 2000
> Date: Wed, 23 Feb 2000 12:59:56 +0100
> From: "Jean-Pierre.EHRMANN"

> Andreas wrote
> > Let P be a point whose the circumcircle of its pedal triangle (pedal circle)
> > touches the nine-point circle. [in Feuerbach's theorem P = Ix].
> > The locus of P is Griffiths' cubic.
> > Now, if P is a point whose the circumcircle of its cevian triangle (cevian
> > circle?) touches the nine-point circle, then the locus is ???
> > (in Feuerbach's theorem P = Gx [=Gergonne's])

> If P is barycentric x | y | z, the two points where the cevian circle of P
> intersects the nine point circle are barycentric
>
> x^2 (z^2 - y^2) [ (c^2 - b^2) y^2 z^2 +b^2 z^2 x^2 - c^2 x^2 y^2 ] , .....
> and
> x [ (c^2 + a^2 - b^2) y - (a^2 + b^2 - c^2) z ] [ (b^2 - c^2) y z + b^2 z
> x - c^2 x y ], .....

So the nine point circle meets every such cevian circle.
Does this characterise the nine-point circle? Or is there another
circle that meets every cevian circle?
--
Barry Wolk
• ... Let me rewrite this so that I can see it: ( xx(yy-zz)[(cc-bb)yyzz + bbzzxx - ccxxyy] : : ) = ( xx(yy-zz)[ccyy(zz-xx) + bbzz(xx-yy)] : : ) = ( ccXY + bbZX
Message 2 of 20 , Mar 2, 2000
On Tue, 29 Feb 2000, Barry Wolk wrote:

> From: Barry Wolk <wolkb@...>
>
> > Date: Wed, 23 Feb 2000 12:59:56 +0100
> > From: "Jean-Pierre.EHRMANN"
>
> > Andreas wrote
> > > Let P be a point whose the circumcircle of its pedal triangle (pedal circle)
> > > touches the nine-point circle. [in Feuerbach's theorem P = Ix].
> > > The locus of P is Griffiths' cubic.
> > > Now, if P is a point whose the circumcircle of its cevian triangle (cevian
> > > circle?) touches the nine-point circle, then the locus is ???
> > > (in Feuerbach's theorem P = Gx [=Gergonne's])
>
> > If P is barycentric x | y | z, the two points where the cevian circle of P
> > intersects the nine point circle are barycentric
> >
> > x^2 (z^2 - y^2) [ (c^2 - b^2) y^2 z^2 +b^2 z^2 x^2 - c^2 x^2 y^2 ] , .....

Let me rewrite this so that I can see it:

( xx(yy-zz)[(cc-bb)yyzz + bbzzxx - ccxxyy] : : )

= ( xx(yy-zz)[ccyy(zz-xx) + bbzz(xx-yy)] : : )

= ( ccXY + bbZX : : ) where X = xx(yy-zz) etc.,

~ ( cc/Z + bb/Y : : )

= ( : cc/zz(xx-yy) + aa/xx(yy-zz) : )

which is the inferior (or subordinate) of ( : bb/yy(zz-xx) : ),
which certainly does lie on the circumcircle, since it's the
conjugal (isogonal conjugate) of ( : yy(zz-xx) : ), which is visibly
on the line at infinity.

> > and
> > x [ (c^2 + a^2 - b^2) y - (a^2 + b^2 - c^2) z ] [ (b^2 - c^2) y z + b^2 z
> > x - c^2 x y ], .....

I rewrite this too:

x[ SB.y - SC.z ]( (bb-cc)yz + bbzx - ccxy ) : :

= x( SB.y - SC.z )( bbz(x+y) - ccy(z+x) ) : :

which should simplify to something of much the same form,
but isn't doing so quite so easily.

Since the first one is invariant under changing the signs of x,y,z,
we get the theorem that the Cevian circles of a harmonically associated
points intersect on the NPC, which is interesting in its own right.

> So the nine point circle meets every such cevian circle.
> Does this characterise the nine-point circle? Or is there another
> circle that meets every cevian circle?

The more interesting remark is that these two points are rational
in aa,bb,cc,x,y,z. Does anyone see an a priori reason why this should
be so?

John Conway
• ... De : John Conway À : Cc : Envoyé : jeudi 2 mars 2000 18:42 Objet : Re:
Message 3 of 20 , Mar 3, 2000
----- Message d'origine -----
De : "John Conway" <conway@...>
À : <Hyacinthos@onelist.com>
Cc : <Hyacinthos@onelist.com>
Envoyé : jeudi 2 mars 2000 18:42
Objet : Re: [EMHL] Re: Proofs of Feuerbach theorem

>Dear all,
>
>
> > Barry Wolk wrote
>
> > So the nine point circle meets every such cevian circle.
> > Does this characterise the nine-point circle? Or is there another
> > circle that meets every cevian circle?
>

John Conway wrote

> The more interesting remark is that these two points are rational
> in aa,bb,cc,x,y,z. Does anyone see an a priori reason why this should
> be so?
>
> John Conway
>
Let us say that a triangle is "self-polar" w.r.t. a conic if the polar of
each vertice is the opposite side - I don't know the English word for this
property -
It is easy to see that if two triangles are self-polar w.r.t. the same
conic, there is a conic going through the six vertices.
Now call W(M) the center of the rectangular hyperbola through A, B, C, M.
We have two self-polar triangles w.r.t. this hyperbola :
- the cevian triangle of M
- U, V, W(M) where U, V are the two infinite points of the circles ( "cyclic
points" ???)
Hence the cevian circle of M goes through W(M) and the cevian circle of M
intersects the NPC at W(M) and W(M°) where M° is the cyclocevian conjugate
of
M. Obviously, those two points are rational in aa, bb, cc, x, y, z.

We have a similar conclusion for the pedal circle of M : he goes through
W(M).
Hence the pedal circle of M intersects the NPC at W(M) and W(M*) where M* is
the isogonal conjugate of M ( the pedal circle is the same one for M and
M* )
Once more, we have two rational points in aa, bb, cc, x, y, z.

Barry Wolk asked the following question :
Is there another circle than NPC that meets every cevian circle?
I can add another one :
Is there another circle than NPC that meets every pedal circle?

For some M, the NPC, the pedal circle and the cevian circle have another
common point.
It is obviously the case for the centroid and the circumcenter - two circles
are the same ones -; in the case of the centroid, the two points are Kiepert
center and Jerabek center.
It is the case for each Fermat point - again, one of the two common points
of the three circles is Kiepert center
The locus of such M is a curve of degree 5

May be, it would be interesting to study the group G generated by isogonal
conjugaison and cyclocevian conjugaison and to try to find the points M such
as the set of W(s(M)) - with s in G - is a finite subset of the NPC.

Sorry for my so bad English language
Friendly from France.
Jean-Pierre
• ... - you happen to have used it! - ... Is M a general point, or some particular one? ... The standard term is circular points at infinity ... Very nice -
Message 4 of 20 , Mar 3, 2000
On Fri, 3 Mar 2000, Jean-Pierre.EHRMANN wrote:

> > > Barry Wolk wrote
> >
> > > So the nine point circle meets every such cevian circle.
> > > Does this characterise the nine-point circle? Or is there another
> > > circle that meets every cevian circle?
>
> John Conway wrote
>
> > The more interesting remark is that these two points are rational
> > in aa,bb,cc,x,y,z. Does anyone see an a priori reason why this should
> > be so?
> >
> > John Conway
> >
> Let us say that a triangle is "self-polar" w.r.t. a conic if the polar of
> each vertice is the opposite side - I don't know the English word for this
> property -

- you happen to have used it! -

> It is easy to see that if two triangles are self-polar w.r.t. the same
> conic, there is a conic going through the six vertices.
> Now call W(M) the center of the rectangular hyperbola through A, B, C, M.

Is M a general point, or some particular one?

> We have two self-polar triangles w.r.t. this hyperbola :
> - the cevian triangle of M
> - U, V, W(M) where U, V are the two infinite points of the circles
> ( "cyclic points" ???)

The standard term is "circular points at infinity"

> Hence the cevian circle of M goes through W(M) and the cevian circle of M
> intersects the NPC at W(M) and W(M�) where M� is the cyclocevian conjugate
> of M. Obviously, those two points are rational in aa, bb, cc, x, y, z.
>
> We have a similar conclusion for the pedal circle of M : he goes through
> W(M).
> Hence the pedal circle of M intersects the NPC at W(M) and W(M*) where M* is
> the isogonal conjugate of M ( the pedal circle is the same one for M and
> M* )
> Once more, we have two rational points in aa, bb, cc, x, y, z.

Very nice - let me summarize to make sure I've got it right -
the joint pedal circle of M and its isogonal conjugate M* passes
through the centers of the rectangular hyperbolae through A,B,C,M
and A,B,C,M*. I now retrospectively recognize this in the coordinates
for the two points.

> Barry Wolk asked the following question :
> Is there another circle than NPC that meets every cevian circle?
> I can add another one :
> Is there another circle than NPC that meets every pedal circle?
>
> For some M, the NPC, the pedal circle and the cevian circle have another
> common point.
> It is obviously the case for the centroid and the circumcenter - two circles
> are the same ones -; in the case of the centroid, the two points are Kiepert
> center and Jerabek center.
> It is the case for each Fermat point - again, one of the two common points
> of the three circles is Kiepert center
> The locus of such M is a curve of degree 5
>
> May be, it would be interesting to study the group G generated by isogonal
> conjugaison and cyclocevian conjugaison and to try to find the points M such
> as the set of W(s(M)) - with s in G - is a finite subset of the NPC.

These remarks are interesting. But let me say that there is some
danger of confusion involving "Cevian circle", one traditional meaning of
which is "circle on a Cevian as diameter", particular cases of which
are the "edge circles", "altitude circles" and "median circles". This
is a useful convention, because there are several theorems about these
circles. The circle you have been discussing is perhaps best called
"the cevian circumcircle" for safety.

John Conway
• First some definitions: ... OK, I will use the term cevian circumcircle of a point P for the circle passing through the three cevian points of P. For P =
Message 5 of 20 , Mar 3, 2000
First some definitions:
John Conway wrote:
> The circle you have been discussing is perhaps best called
> "the cevian circumcircle" for safety.

OK, I will use the term "cevian circumcircle" of a point P
for the circle passing through the three cevian points of P.

For P = (x:y:z) in barycentrics,
iso(P) = (1/x : 1/y : 1/z) = the isotomic conjugate of P,
sub(P) = (y+z : z+x : x+y),
super(P) = (y+z-x : z+x-y : x+y-z)

Now a few results:
The 9PC is the cevian circumcircle of G, and also that of the
orthocenter H. Does this generalize? Well, it is easy to show that,
for any two points P and Q, the cevian points of P and the cevian
points of Q all lie on a conic. And I have just proved that P and Q
have the same cevian circumcircle iff sub(iso(P)) and sub(iso(Q))
are isogonal conjugates. The hardest part of this was finding the
result which was to be proved -- the proof itself wasn't that bad,
just some fiddling with the equations.

This is related to what Jean-Pierre.EHRMANN wrote:
> May be, it would be interesting to study the group G generated
> by isogonal conjugaison and cyclocevian conjugaison and to try
> to find the points M such as the set of W(s(M)) - with s in G -
> is a finite subset of the NPC.

I didn't know there was a term "cyclocevian conjugate". So the
cyclocevian conjugate of P is iso(super(con(sub(iso(P))))).
:-)

Finally, a note for Steve's thread "Unusual concurrences and cublcs"
The twelve points of any desmic configuration always lie on a cubic.
Specifically, the points
(U+u : V+v : W+w) (1:0:0) (0:1:0 (0:0:1)
(U:V:W) (U:v:w) (u:V:w) (u:v:W)
(u:v:w) (u:V:W) (U:v:W) (U:V:w)

all lie on the cubic

x(u+U)(wWyy-vVzz) + y(v+V)(uUzz-wWxx) + z(w+W)(vVxx-uUyy) = 0
--
Barry Wolk <wolkb AT cc.umanitoba.ca>
• ... De : John Conway À : Cc : Hyacinthe Envoyé : vendredi 3 mars 2000 17:10
Message 6 of 20 , Mar 5, 2000
----- Message d'origine -----
De : "John Conway" <conway@...>
À : <Hyacinthos@onelist.com>
Cc : "Hyacinthe" <Hyacinthos@onelist.com>
Envoyé : vendredi 3 mars 2000 17:10
Objet : Re: [EMHL] Re: Proofs of Feuerbach theorem

> From: John Conway <conway@...>
>
> Dear all, John Conway wrote
>
> On Fri, 3 Mar 2000, JP Ehrmann wrote:

Now call W(M) the center of the rectangular hyperbola through A, B, C, M
>
> Is M a general point, or some particular one?
>
> >

I mean any point : W(M) is the midpoint of R*H,
with R = infinite point of OM*.

Looking at this configuration, I got this - may be it is well known, I don't
know; in any case, I think it is funny - :
Let P(A, B, C, D, E) the following property :
The perpendicular lines from E to AD, BD, CD meet respectively BC, CA, AB in
three points lying on a same line.
Then P(A,B,C,D,E) is true iff the rectangular hyperbola through A,B,C,D
passes through E.
Hence P is invariant after any permutation.

Friendly from France. JPE
• ... De : Barry Wolk À : Envoyé : vendredi 3 mars 2000 22:20 Objet : [EMHL] Re: Proofs of Feuerbach theorem
Message 7 of 20 , Mar 7, 2000
----- Message d'origine -----
De : "Barry Wolk" <wolkb@...>
À : <Hyacinthos@onelist.com>
Envoyé : vendredi 3 mars 2000 22:20
Objet : [EMHL] Re: Proofs of Feuerbach theorem

dear all,
Barry Wolk wrote
> Finally, a note for Steve's thread "Unusual concurrences and cublcs"
> The twelve points of any desmic configuration always lie on a cubic.
> Specifically, the points
> (U+u : V+v : W+w) (1:0:0) (0:1:0 (0:0:1)
> (U:V:W) (U:v:w) (u:V:w) (u:v:W)
> (u:v:w) (u:V:W) (U:v:W) (U:V:w)
>
> all lie on the cubic
>
> x(u+U)(wWyy-vVzz) + y(v+V)(uUzz-wWxx) + z(w+W)(vVxx-uUyy) = 0
>
If I well understand what you mean by desmic configuration, the
three-dimensionnal configuration
[0,0,0,1] [1,0,0,0] [0,1,0,0] [0,0,1,0]
[x,y,z,t] [-x,y,z,-t ] [-x,y,-z,t] [-x,-y,z,t]
[x,y,z,-t] [-x,y,z,t] [x,-y,z,t] [x,y,-z,t]
should have the same properties :
12 points, 16 lines with 3 of the points on each one, each point lying on 4
lines.
Is that right?
Hence any perspective of this configuration on a plane should be a desmic
configuration.
For instance we can start from the vertices of a tetrahedron and the centers
of the spheres tangent to the four faces - with the areas of the faces for
x,y,z,t -
Hence, the vertices of a face F of a tetrahedron, the projection of the
fourth vertice and the points where the spheres tangent to the four faces
touch F should be a desmic configuration.
Friendly from France.
Jean-Pierre
• ... Yes, that s what desmic means, 12 points, 16 lines, etc. But John Conway once remarked that the planar desmic configurations are more general than just the
Message 8 of 20 , Mar 9, 2000
Jean-Pierre.EHRMANN wrote:

> Barry Wolk wrote
> > Finally, a note for Steve's thread "Unusual concurrences and cublcs"
> > The twelve points of any desmic configuration always lie on a cubic.
> > Specifically, the points
> > (U+u : V+v : W+w) (1:0:0) (0:1:0) (0:0:1)
> > (U:V:W) (U:v:w) (u:V:w) (u:v:W)
> > (u:v:w) (u:V:W) (U:v:W) (U:V:w)
> >
> > all lie on the cubic
> >
> > x(u+U)(wWyy-vVzz) + y(v+V)(uUzz-wWxx) + z(w+W)(vVxx-uUyy) = 0
>
> If I well understand what you mean by desmic configuration, the
> three-dimensionnal configuration
> [0,0,0,1] [1,0,0,0] [0,1,0,0] [0,0,1,0]
> [x,y,z,t] [-x,y,z,-t ] [-x,y,-z,t] [-x,-y,z,t]
> [x,y,z,-t] [-x,y,z,t] [x,-y,z,t] [x,y,-z,t]
> should have the same properties :
> 12 points, 16 lines with 3 of the points on each one, each point
> lying on 4 lines. Is that right?
> Hence any perspective of this configuration on a plane should be
> a desmic configuration.

Yes, that's what desmic means, 12 points, 16 lines, etc. But John Conway
once remarked that the planar desmic configurations are more general
than just the projections of that 3D case. The general planar example
can be assigned coordinates as above, namely (U:V:W), (U:v:w), etc

The above cubic can be written as the vanishing of the determinant
| U+u V+v W+w |
| x y z |
| Uu/x Vv/y Ww/z |

We have three quadrangles (Io,Ia,Ib,Ic), (IIo,IIa,etc), (IIIo,etc)
with labels and coordinates as follows:

x | o a b c
______|____________________________________________
Ix | (U+u:V+v:W+w) (1:0:0) (0:1:0) (0:0:1)
IIx | (U:V:W) (U:v:w) (u:V:w) (u:v:W)
IIIx | (u:v:w) (u:V:W) (U:v:W) (U:V:w)

The diagonal points of the first quadrangle are
(0:V+v:W+w), (U+u:0:W+w), (U+u:V+v:0).
And I noticed that these diagonal points also lie on this cubic.
It follows from the symmetry of the situation that the diagonal
points of each of the three quadrangles all lie on this cubic.
I verified this, and then found that there are enough collinearities
among these 9 diagonal points to create another desmic configuration,
with all 24 points lying on the same cubic curve.

Here are the results of some computer calculations.
Define the following expressions:
E1 = (UVW+Uvw-uVw-uvW) /(U-u)
e1 = (uvw+uVW-UvW-UVw) /(u-U)
E2 = (UVW-Uvw+uVw-uvW) /(V-v)
e2 = (uvw-uVW+UvW-UVw) /(v-V)
E3 = (UVW-Uvw-uVw+uvW) /(W-w)
e3 = (uvw-uVW-UvW+UVw) /(w-W)

Note en is obtained from En by swapping U with u, swapping V with v,
and swapping W with w.

Then the new desmic configuration is

x o a b c
I'x [Uu/(U+u):Vv/(V+v):Ww/(W+w)] [0:V+v:W+w] [U+u:0:W+w] [U+u:V+v:0]
II'x [UuE1:VvE2:WwE3] [2Uu:E3:E2] [E3:2Vv:E1] [E2:E1:2Ww]
III'x [Uue1:Vve2:Wwe3] [2Uu:e3:e2] [e3:2Vv:e1] [e2:e1:2Ww]

The last three columns are the diagonal points of the corresponding
quadrangles. For example, line IIo IIb meets line IIa IIc at II'b.
And these 12 points lie on the same cubic curve as the original 12 points.

In brief, the 9 diagonal points of 3 desmically-related quadrangles are part
of another desmic configuration, and all 24 points lie on a cubic curve.

[And after doing all this the hard way, I realized that this is
a consequence of the group structure of the cubic curve that
passes through the original 12 points.]
--
Barry Wolk <wolkb AT cc.umanitoba.ca>
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