----- Message d'origine -----

De : "Steve Sigur" <

ssigur@...>

� : <

Hyacinthos@onelist.com>

Envoy� : mardi 1 f�vrier 2000 05:53

Objet : Re: [EMHL] More on the Jerabek points

Dear Steve and all,

Steve wrote

Jean- Pierre wrote

"There is an easy way to generalize the Brocard points and the first

> >Brocard triangle :

> >Take any point M on the line GK; the parallel lines from M to BC, CA, AB

> >intersect the medians AA1, BB1, CC1 at Pa, Pb, Pc.

> >Let Ma, Mb, Mc the reflections of M with respect to Pa, Pb, Pc.

> >Then the triangle MaMbMc is triple perspective and similar to ABC. "

> >> I think this result is significant because the line GK is a strong

line,

> and I suspect from their construction that Ma, Mb, and Mc are strong

> points.

>

> In playing with this on GSP I have found that as M moves, Ma Mb and Mc

> all move on a straight line through G and the perspectors dance some more

> complicated curve that goes twice through each vertex and 3 times through

> G. I do not recognize this curve, but you might.

>

> In fact, an homothecy with center G maps the first Brocard triangle to

MaMbMc and your curve is the union of three hyperbolas through A, B, C, G,

one of them being Kiepert hyperbola.

I know some tiny results, almost nothing; for instance :

- if you reflect ABC with respect to the line

U X + V Y + W Z = 0, the new triangle is triple perspective with ABC iff

a^4 U^4 - 2 b^2 c^2 V^2 W^2 + 4 (SA^2 - (a^2 - b^2)(a^2 - c^2)] U^2 V W +

...(similar terms) = 0, but it seems to be the tangential equation of a

quite complicated curve.

- if you rotate ABC around a point of the circumcircle, for particular

values of the angle, the new triangle is triple perspective with ABC

Friendly from France.

Jean-Pierre