## Extra Summer Laboratory of triangles

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• Friends of Hyacinthos: Four problems are offered  in  http://personal.us.es/rbarroso/trianguloscabri/ One of the inradius and circumradius, of Vcente Vicario
Message 1 of 10 , Jun 30, 2010
Friends of Hyacinthos:

Four problems are offered

in

http://personal.us.es/rbarroso/trianguloscabri/

another on the
equilateral, proposed for Ricard Peiró

and two of Paul Halmos
Greetings from
Sevilla

Ricardo
Barroso

[Non-text portions of this message have been removed]
• Dear Hauke, Barry and Jean-Pierre! ... I don t agree with you. Your circles are member of the pencil with Poncelet points (circles with radius 0) A and the
Message 2 of 10 , Jul 2, 2010
Dear Hauke, Barry and Jean-Pierre!

> Some computation shows that for any circle that meets the 2 sides AB and AC harmonically, its center lies on the perpendicular bisector of BC.

I don't agree with you. Your circles are member of the pencil with Poncelet points (circles with radius 0) A and the projection of A upon BC.

The conic meets AB harmonically iff A lyes on the polar of B. Thus triangle ABC is autopolar wrt Hauke's conic. But in this case all six common points of this conic and the sidelines can't be real.

Sincerely Alexey

[Non-text portions of this message have been removed]
• ... I finally realized that I used an incorrect formula for harmonic conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct (BP/PC)/(BQ/QC)=-1 when P,Q
Message 3 of 10 , Jul 2, 2010
> From: jpehrmfr <jean-pierre.ehrmann@...>
>
> Dear Barry
> [Barry Wolk]
> > Some computation shows that for any circle that meets
> the 2 sides AB and AC harmonically, its center lies on the
> perpendicular bisector of BC.
>
> I don't agree with you. Your circles are member of the
> pencil with Poncelet points (circles with radius 0) A and
> the projection of A upon BC.
> Friendly. Jean-Pierre

I finally realized that I used an incorrect formula for harmonic conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results apply to this strange kind of "conjugates," but not to harmonics.
--
Barry Wolk
• Dear Hyacinthians, Finally catching up on my mail. At the risk of having missed part of the discussion, I only partially agree with Alexey (if I understand him
Message 4 of 10 , Jul 19, 2010
Dear Hyacinthians,

Finally catching up on my mail. At the risk of having missed part of the
discussion, I only partially agree with Alexey (if I understand him
correctly). Using his observation that ABC is autopolar with respect to
Hauke's conic, it is a straightforward exercise in barycentric calculus
to show that any circle with the property that Hauke is looking for
always has its center at the orthocenter of ABC. Also, with a little
more work it can be shown that the radius of such a circle has to be
equal to 2*R*sqrt(-cos(aA)cos(aB)cos(aC)), with R the radius of the
circumcircle of ABC and aA, aB, aC the angles at A, B, C. It follows
that there is no such (real) circle for acute triangles and exactly one
for obtuse triangles.

Eisso

On 7/2/10 3:42 AM, Alexey Zaslavsky wrote:
>
> Dear Hauke, Barry and Jean-Pierre!
>
> > Some computation shows that for any circle that meets the 2 sides AB
> and AC harmonically, its center lies on the perpendicular bisector of BC.
>
> I don't agree with you. Your circles are member of the pencil with
> Poncelet points (circles with radius 0) A and the projection of A upon BC.
>
> The conic meets AB harmonically iff A lyes on the polar of B. Thus
> triangle ABC is autopolar wrt Hauke's conic. But in this case all six
> common points of this conic and the sidelines can't be real.
>
> Sincerely Alexey
>
>

--

========================================
Eisso J. Atzema, Ph.D.
Department of Mathematics& Statistics
University of Maine
Orono, ME 04469
Tel.: (207) 581-3928 (office)
(207) 866-3871 (home)
Fax.: (207) 581-3902
E-mail: atzema@...
========================================

[Non-text portions of this message have been removed]
• I just remember the Faure theorem. A triangle ABC is autopolar wrt the conic Gamma iff the ABC-circumcircle is orthogonal to the orthoptic circle of Gamma.
Message 5 of 10 , Jul 29, 2010
I just remember the Faure theorem.
A triangle ABC is autopolar wrt the conic Gamma iff the ABC-circumcircle is
orthogonal to the orthoptic circle of Gamma.
Friendly
Francois

On Fri, Jul 2, 2010 at 8:15 PM, Barry Wolk <wolkbarry@...> wrote:

>
>
> >
>
> >
> > Dear Barry
> > [Barry Wolk]
> > > Some computation shows that for any circle that meets
> > the 2 sides AB and AC harmonically, its center lies on the
> > perpendicular bisector of BC.
> >
> > I don't agree with you. Your circles are member of the
> > pencil with Poncelet points (circles with radius 0) A and
> > the projection of A upon BC.
> > Friendly. Jean-Pierre
>
> I finally realized that I used an incorrect formula for harmonic
> conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct
> (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results
> apply to this strange kind of "conjugates," but not to harmonics.
> --
> Barry Wolk
>
>
>

[Non-text portions of this message have been removed]
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