> Date: Wed, 23 Feb 2000 12:59:56 +0100

So the nine point circle meets every such cevian circle.

> 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 ], .....

Does this characterise the nine-point circle? Or is there another

circle that meets every cevian circle?

--

Barry Wolk- Jean-Pierre.EHRMANN wrote:

> Barry Wolk wrote

Yes, that's what desmic means, 12 points, 16 lines, etc. But John Conway

> > 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.

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>