## Re: More on the Jerabek points

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• ... De : Steve Sigur À : Envoyé : mardi 1 février 2000 05:53 Objet : Re: [EMHL] More on the Jerabek points
Message 1 of 7 , Feb 1, 2000
----- 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
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